MLR2-1 str. 1
Translation from Polish: Iwona Turnau, prof. Stefan Turnau
Book design: Iwona Duczmal
Ilustrations: Bartłomiej Brosz
Computer generated graphics: Leszek Jakubowski
Photography: Archiwum GWO, Fotolia, Shutterstock
Typesetting (TEX): Joanna Szyller
ISBN 978-83-8118-377-2
Publisher: Gdańskie Wydawnictwo Oświatowe, 80–309 Gdańsk, al. Grunwaldzka 413
This publication is subject to the protection provided by the provisions of the Act
of 4 February 1994 on copyright and neighbouring rights. Any copy or reproduction of a part or of the whole publication constitutes an unauthorized infringement
of the rights of the author or publisher, unless it is performed in accordance with
the provisions of the aforementioned act.
MLR2-1 str. 2
Table of Contents
Polynomials
Examples of polynomials
.....................................................................................
Decomposition of a polynomial into factors
Polynomial equations
10
.........................................................................................
15
Division of polynomials
Remainder Theorem
6
.....................................................
.....................................................................................
21
...........................................................................................
27
Polynomial equations (continued)
.....................................................................
30
Figures on the plane. Part 1
Angles. Angles in triangles and quadrilaterals
Basic properties of triangles
..................................................
39
..............................................................................
45
Pythagoras’ theorem and inverse of Pythagoras’ theorem
Properties of triangles (continued)
................................
47
....................................................................
53
...............................................................................
62
.........................................................................................
70
Properties of quadrilaterals
Functions
Polynomial functions
Polynomial inequalities
......................................................................................
Polynomial functions (continued)
......................................................................
Polynomial inequalities (continued)
...................................................................
75
79
83
Exponential and logarithmic functions
.............................................................
85
Exponential and logarithmic equations
.............................................................
91
Applications of exponential and logarithmic functions
....................................
97
Figures on the plane. Part 2
Area of a disc. Length of a circle
.....................................................................
Properties of central angles and inscribed angles
Lines and circles
...........................................
108
...............................................................................................
112
Circle circumscribed about a polygon
Circle inscribed in a polygon
.............................................................
117
...........................................................................
122
Properties of polygons. Regular polygons
MLR2-1 str. 3
104
.......................................................
126
MLR2-1 str. 4
Polynomials
Do you know how many donuts the pastry chef needs
to form a five-layer pyramid, such as in the photo?
How many donuts would he need if the layers were 12?
This problem can be easily resolved using a certain algebraic expression.
Examples of polynomials
Polynomial equations
Remainder Theorem
MLR2-1 str. 5
Decomposition of a polynomial into factors
Division of polynomials
Polynomial equations (continued)
EXAMPLES OF POLYNOMIALS
Examples of monomials:
4x16
2
y
3
−3m5
√
−3 2t 101
Each of the expressions in the adjacent box
is the product of a number and the power of
one variable with a natural exponent. We call
these expressions monomials.
Note. We also call monomials expressions with
3
more variables, e.g. 3x2 y, ab3 , 5 m5 n6 are monomials. However, we will not deal with them in
this chapter.
A monomial of degree n ≥ 1 and variable
x is an expression that can be transformed
into the form:
axn , where a ∈ ’, a 6= 0, n ∈ Ž+ .
Note, that any real number other than zero is a zero-degree monomial.
We assume that the number 0 is also a monomial and we call it a zero
monomial. The degree of such monomial is not specified.
Monomials and their sums are called polynomials.
A polynomial of degree n and variable x
is an expression that can be converted to
the form:
an xn + an−1 xn−1 + ... + a2 x2 + a1 x + a0 ,
Examples of polynomials:
4x5 + 11x3 + 7
−8a3 − 2 a5
3
where the coefficients an , an−1 , ..., a2 , a1 , a0
are real numbers, an 6= 0 and n ∈ Ž.
The coefficient a0 of a polynomial is called
the free term.
0,04u8
√
3t 7 − 2 6t 5 + t 4 − 9
(2x − 1)2
2 + m + 2m2
3
EXERCISE A Specify the degree of each polynomial given in the box.
6
6
POLYNOMIALS
MLR2-1 str. 6
Examples of binomials:
2x − 1
a3 + 2
√
3x5 − 2x
3 7
t
4
+ t 12
A polynomial that can be written in the
form of the sum of two non-zero monomials of different degrees is called a binomial, and the sum of three monomials (of
different degrees) is called a trinomial.
A trinomial, which is a second-degree polynomial, is called a quadratic trinomial.
3x2 − 2x + 3
EXERCISE B a) List the coefficients at the
highest power of each of the binomials and
trinomials in the box.
2 8
y
3
b) Give an example of quadratic trinomial
with integer coefficients.
Examples of trinomials:
+ y4 + 1
m10 − 2m7 + m6
√ 5
7x − x − x6
c) Is the expression (2x − 3)2 a quadratic trinomial?
Polynomials can be added, subtracted and multiplied. Performing this type
of operation, we get a new polynomial, which is worth ordering, i.e. reducing similar terms and ordering the monomials from highest to lowest
degree.
Write in the simplest form:
EXAMPLE 1
a) the sum of polynomials:
(7 − 5x 5 − 3x 2 ) + (3x 2 − 4x − x 5 ) = 7 − 5x 5 − 3x 2 + 3x 2 − 4x − x 5 = −6x 5 − 4x + 7
b) the difference of polynomials:
(−8p 4 − 2p 6 + 3) − (3 − 5p + 2p 6 ) = −8p 4 − 2p 6 + 3 − 3 + 5p − 2p 6 = −4p 6 − 8p 4 + 5p
c) the product of polynomials:
(3−2t 5 +t)(−5t 2 −t) = −15t 2 −3t +10t 7 +2t 6 −5t 3 −t 2 = 10t 7 + 2t 6 − 5t 3 − 16t 2 − 3t
PROBLEM
5
Put in order the polynomial:
2
a) (4x + 3x − 2x + 5) + (2x5 − 3x2 + 3x − 5)
b) (9a6 − 5a3 ) − (9a6 + 5a4 − 4a3 + 2)
c) (3z 5 − 2z)(z 2 − 2z + 3)
The variable of a polynomial can of course be any letter. We will sometimes
denote polynomials with variable x in short W (x), V (x), P (x).
EXAMPLES OF POLYNOMIALS
MLR2-1 str. 7
7
The value of a polynomial for a given number is obtained by substituting
this number for the variable in the polynomial. For a given polynomial
W (x)), its value for x = 5 will be written as W (5).
EXAMPLE 2
2
Calculate the value of the given polynomial for x = 5 .
a) W (x) = (3x 7 − 9x 3 − 4x) − (3x 7 − 8x 3 + x)
W (x) = 3x 7 − 9x 3 − 4x − 3x 7 + 8x 3 − x = −x 3 − 5x
3
2
2
2
8
8
W 5 = − 5 − 5 × 5 = − 125 − 2 = −2 125
b) V (x) = (25x 2 − 5x + 3)(1 − x)
2
2
2
2
2
V 5 = 25 × 5 − 5 × 5 + 3 1 − 5 = 3
PROBLEM
To simplify calculations,
we first put in order the
polynomial W (x).
We do not order the polynomial V (x) because it
does not simplify calculations.
Calculate the value of the given polynomial for x = 1 .
3
a) W (x) = (8x5 + 6x2 + 3) + (−8x5 − 5x2 + x − 3)
b) V (x) = (3x − 2)(x3 − 9x2 + 1)
Two polynomials of the same variable are equal when they are of the
same degree, and after ordering each of them, the coefficients at the same
powers are equal.
EXAMPLE 3
Let’s consider polynomials:
U(x) = ax 2 + bx
V (x) = 2x 3 − 11x 2 + 12x
W (x) = x − 3
For what values of the coefficients a and b the polynomial U(x) − V (x) ) is equal
to the polynomial U(x) × W (x)?
U(x) − V (x) = (ax 2 + bx) − (2x 3 − 11x 2 + 12x) = ax 2 + bx − 2x 3 + 11x 2 − 12x =
= −2x 3 + (a + 11)x 2 + (b − 12)x
U(x) × W (x) = (ax 2 + bx)(x − 3) = ax 3 − 3ax 2 + bx 2 − 3bx = ax 3 + (b − 3a)x 2 − 3bx
U(x) − V (x) = −2x 3 + (a + 11)x 2 + (b − 12)x
U(x) × W (x) = ax 3 + (b − 3a)x 2 − 3bx



 −2 = a
a + 11 = b − 3a



b − 12 = −3b
Therefrom: a = −2 and b = 3
We compare the coefficients of both polynomials at the respective powers of the variable; we solve a system of equations.
The numbers a = −2 and b = 3 satisfy each of
the three equations of the system.
PROBLEM
For what values of the a and b coefficients the product of the polynomials U(x) = 2x − a and V (x) = x3 − x is equal to the polynomial W (x) = bx4 − 5x3 − 2x2 + ax?
8
POLYNOMIALS
MLR2-1 str. 8
PROBLEMS
1. Indicate the polynomials in the box and specify the degree of each.
a)
b)
c)
7x5 − 5x7
2
6u3 − 11u−2 + 4
√ 105 z 5
3z
+
1
x3 − 5x2 + 4
0,2t + 6t 3 − 1,4t 10
√
−2x6 − 5 x + 4
−3w 7
1
1
5w 3 + 4w
82
3
2. Present the given expression in the form axn .
a)
√
b) x2 + x2 2
3 2
− 1 x2 × − 2 x
2
3
2
2
d) 5x − (3x)
3
c) 3x × x2 − 1 x5
2
5
2
3. Present the result of the operation in the form of an ordered polynomial.
4x − 8x3 − 2x5 − 3x + 2 − 8x3
a)
f)
2
g) (x − 6)(2x + 3) + 3x 4 − 2x2
5
9 + 7x3 )
4
10
4x(−2x
h)
− 20x − 8x
× (1 − 2x)
2
5
−3x5 + 5x7 − 2 + 3 − 2x − 7x
b)
2
2
5
3
c) 6x + 4x − 2x − 2x(2x + x − 1)
2
5
3
7
5
6
7
d) x(x − x ) − x (x − x − 1)
3
2
i)
4
3
4
e) −x(x − 8x − 5) − 4x (3x − 2)
j)
6
5x3 − 2 x3 + 1 − 6x3 2 − x3
3
12
(3 − 4x3 )(5x2 + x) − 6x5 − 5x3 − 5x5 − x3 2
h
4
6
2 − 3x2 × x − 7x − 2
i
3
4. Let P
denote polynomial −4x+5, Q — polynomial x2 −3x+1, and R — polynomial
2x − 1. Perform the operations.
3
b) 4Q − 3P + 1 R
a) P − (Q + R)
2
c) R × (P + Q)
5. Give examples of two fourth-degree polynomials of which:
a) the sum is a third-degree monomial,
b) the product is a binomial.
6. Calculate the value of the polynomial for the given x value.
a) 0,32x3 − 0,42x − 1,32x3 + 1,42x, gdy x = −10
√ √
b) 3 2 x4 − 3 x3 + 2x2 + 2 5 + 1 x4 + 1 x3 − 3x2 , gdy x = 5
3
5
5
3
5
4
c) −2,3x + 0,6x − 1,7x − 1,7x5 + 1,6x4 − 0,7x , gdy x = −0,1
3
d) −9 3 x3 + 3 5 x2 + 1 3 x3 + x − 2 x2 (4x − 1) , gdy x = 1
7
7
7
7
2
3
3
2
e) 0,6x − 1,4 + 7,4x + 3,4x + 8,6x + 1,15 (3 − 8x) , gdy x = 0,25
EXAMPLES OF POLYNOMIALS
MLR2-1 str. 9
9
7. The confectioner arranges donuts in pyramids, as
in the photo. Number of donuts in the pyramid with
n layers is:
L(n) = 31 n3 + 12 n2 + 16 n
a) Calculate how many donuts a pyramid sized 12
layers is made of.
b) Justify that L(n + 1) − L(n) = (n + 1)2 .
DECOMPOSITION OF A POLYNOMIAL
INTO FACTORS
It is known that as a result of multiplication of polynomials we get a
certain polynomial. Sometimes the reverse operation can be performed —
decomposing a polynomial into factors, i.e. presenting it as the product of
other polynomials. We call it factorization.
EXERCISE A
5
What monomials should be written in place of the boxes?
3
a) x + 2x =
(x2 +
)
b) 6x4 − 9x = 3x(
−
)
Decompose the polynomial into factors of at most second degree.
EXAMPLE 1
a) 6x 3 − 3x 2 + 10x − 5 =
2
= 3x (2x − 1) + 5(2x − 1) =
= (2x − 1)(3x 2 + 5)
b) 5x 4 − 20x 3 − x 2 + 4x =
= 5x 3 (x − 4) − x(x − 4) =
In each of the underlined binomials we take
out the common factor before the parenthesis, so as to obtain the sum of polynomials
that have a common factor. Then this common factor, i.e. 2x − 1, is taken out.
In each of the underlined binomials, we take
out the common factor, so as to obtain the
sum of polynomials that have a common
factor.
= (x − 4)(5x 3 − x) =
The x − 4 binomial is taken out.
= x(x − 4)(5x 2 − 1)
In the underlined binomial, we take common
factor x out.
PROBLEM
3
Factorize the polynomial.
2
a) 10x − 4x + 15x − 6
10
b) 6x4 − 15x3 − 2x2 + 5x
POLYNOMIALS
MLR2-1 str. 10
Some quadratic trinomials can be broken down into factors using the formulas we recall below.
The equation ax2 + bx + c = 0, where a 6= 0, can be represented in the
product form, when ∆ ≥ 0 (∆ = b2 − 4ac).
If ∆ > 0, then the equation ax2 + bx + c = 0 has two solutions:
x1 =
√
−b − ∆
2a
x2 =
√
−b + ∆
2a
and equality occurs:
ax 2 + bx + c = a(x − x1 )(x − x2 )
If ∆ = 0, then the equation ax2 + bx + c = 0 has one solution:
x0 =
−b
2a
and equality occurs:
ax 2 + bx + c = a(x − x0 )2
If ∆ < 0, then the equation ax2 + bx + c = 0 has no solutions and
the polynomial ax2 + bx + c cannot be broken down into first-degree
factors.
Decompose the polynomial W (x) into factors of the lowest degree.
EXAMPLE 2
a) W (x) = −10x 3 + 25x 2 + 60x = −5x(2x 2 − 5x − 12)
∆ = (−5)2 − 4 × 2 × (−12) = 121
√
5 − 121
3
x1 =
=−
2×2
2
√
5 + 121
=4
x2 =
2×2
We calculate the determinant ∆ of the trinomial
2x 2 − 5x − 12. Because ∆ > 0, the trinomial is
broken down into first-degree factors.
3
3
W (x) = −5x × 2 x − −
(x − 4) = −10x x +
(x − 4)
2
2
b) W (x) = x 5 − 3x 4 + 4x 3 =
3
We calculate the determinant ∆ of the trinomial
x 2 − 3x + 4. Because ∆ < 0, this trinomial does
not break down into first-degree factors.
2
= x (x − 3x + 4)
∆ = 9 − 4 × 1 × 4 = −7 < 0
PROBLEM
Break down the polynomial W (x) into factors of possibly lowest degree.
5
a) W (x) = −4x + 14x4 − 6x3
b) W (x) = −x4 + 4x3 − 7x2
DECOMPOSITION OF A POLYNOMIAL INTO FACTORS
MLR2-1 str. 11
11
Below we recall the abridged multiplication formulas. They can be used to
factorizing polynomials.
Abridged multiplication formulas
(a + b)2 = a 2 + 2ab + b 2
(a + b)3 = a 3 + 3a 2 b + 3ab 2 + b 3
(a − b)2 = a 2 − 2ab + b 2
(a − b)3 = a 3 − 3a 2 b + 3ab 2 − b 3
a 2 − b 2 = (a − b)(a + b)
a 3 − b 3 = (a − b)(a 2 + ab + b 2 )
a 3 + b 3 = (a + b)(a 2 − ab + b 2 )
a n − b n = (a − b)(a n−1 + a n−2 b + a n−3 b 2 + . . . + ab n−2 + b n−1 )
EXERCISE B Justify that the formulas for the difference of squares and the difference of cubes result from the formula for factorizing the difference an − bn .
Decompose the polynomial into factors of the lowest degree pos-
EXAMPLE 3
sible.
a) x 4 − 25 = (x 2 − 5)(x 2 + 5) =
√
√
= (x − 5)(x + 5)(x 2 + 5)
We use the formula
a 2 − b 2 = (a − b)(a + b) twice.
b) x 5 + x 4 + x 3 − 8x 2 − 8x − 8 =
We take out the common factor in each of
the underlined trinomials.
= x 3 (x 2 + x + 1) − 8(x 2 + x + 1) =
We take out the common factor x 2 + x + 1.
= (x 2 + x + 1)(x 3 − 8) =
= (x 2 + x + 1)(x − 2)(x 2 + 2x + 4)
∆1 = −3 < 0
We check whether the second-degree factors obtained can be broken down into firstdegree factors.
∆2 = −12 < 0
c) x 6 + 2x 3 + 1 = (x 3 )2 + 2x 3 + 1 =
3
2
2
2
= (x + 1) = [(x + 1)(x − x + 1)] =
= (x + 1)2 (x 2 − x + 1)2
∆ = −3 < 0
PROBLEM
We use the formula
a 3 − b 3 = (a − b)(a 2 + ab + b 2 ).
We use the formula a 2 +2ab+b 2 = (a+b)2 , and
then the formula a 3 + b 3 = (a + b)(a 2 − ab + b 2 ).
We check whether the second-degree factor obtained can be broken down into firstdegree factors.
Factorize the polynomial.
a) 81x4 − 1
12
b) 8x5 + 8x3 − x2 − 1
c) x4 − 6x2 + 9
POLYNOMIALS
MLR2-1 str. 12
Sometimes, to decompose a polynomial into factors, you have to be smart
and use unusual tricks, e.g. present a monomial as the sum of two monomials or add and subtract the same monomial.
EXAMPLE 4
Decompose polynomials into factors.
a) 2x 4 + x 3 + 3x 2 + x + 1 =
We replace the monomial 3x 2 with the sum
x 2 + 2x 2 .
= 2x 4 + x 3 + x 2 + 2x 2 + x + 1 =
2
2
2
= x (2x + x + 1) + 2x + x + 1 =
In the selected trinomial, we take out the
common factor.
We take the common factor (2x 2 + x + 1) out.
= (2x 2 + x + 1)(x 2 + 1)
b) 4x 3 − 5x + 1 =
We replace the monomial −5x by the sum
−4x − x.
= 4x 3 − 4x − x + 1 =
In the marked binomials, we take common
factor out.
2
= 4x(x − 1) − (x − 1) =
We use the formula a 2 − b 2 = (a − b)(a + b).
= 4x(x − 1)(x + 1) − (x − 1) =
h
i
= (x − 1) 4x(x + 1) − 1 =
We take out the common factor (binomial
x − 1).
= (x − 1)(4x 2 + 4x − 1)
c) x 4 + 4 =
We add and subtract 4x 2 monomial.
= x 4 + 4x 2 + 4 − 4x 2 =
We use the formula a 2 + 2ab + b 2 = (a + b)2 .
= (x 2 + 2)2 − 4x 2 =
We use the formula a 2 − b 2 = (a − b)(a + b).
2
2
= (x + 2 − 2x)(x + 2 + 2x) =
= (x 2 − 2x + 2)(x 2 + 2x + 2)
PROBLEM
Factorize the polynomial.
a) 2x4 + x3 + 5x2 + x + 3
b) 3x3 − 4x − 1
c) x4 + 1
In the examples discussed so far, the factors appearing in the polynomial’s
decomposition did not have a degree higher than 2. It turns out that each
polynomial can be broken down into factors of at most second degree.
DECOMPOSITION OF A POLYNOMIAL INTO FACTORS
MLR2-1 str. 13
13
Good to know!
Already in the eighteenth century, the statement was known:
Each polynomial can be broken down into factors of at most second degree.
Carl Friedrich Gauss (1777–1855) gave the first proof of this claim in his
doctoral dissertation. He was only 22 years old. Gauss is considered, along
with Archimedes and Newton, to be one of the greatest mathematicians
of the world (he was called the prince of mathematicians). He dealt with
almost all branches of mathematics, as well as physics and astronomy.
PROBLEMS
1. Take out the common factor
a) x5 − x3
c) 4x7 + 8x6
e) −9x6 + 18x4 − 12x3
b) x4 − x3 + x2
d) 15x5 − 20x3 + 5x2
f) 14x8 − 21x7 − 28x5
2. Present as the product of two polynomials.
a) x3 + 4x2 + x + 4
c) x3 − 1 x2 + x − 1
e) 3x6 − 4x5 − 3x3 + 4x2
b) 6x3 − 5x2 + 6x − 5
d) 2x4 + 5x3 − 2x2 − 5x
f) −7x3 − 6x2 + 7x + 6
2
2
3. Factorize the polynomial.
a) x3 + 5x2 + 3x + 15
d) −15x3 + 6x2 − 5x + 2
√
√
e) 5x3 − 7x2 + 10x − 2 7
√ 3 √ 2
√
√
f)
2x − 3x + 5 2x − 5 3
b) 2x3 − 3x2 + 6x − 9
c) 2x3 − 3x2 + 4x − 6
4. Factorize the polynomial.
a) x2 − 16
e) x2 − 6x + 9
b) 4x2 − 5
f) 1 x2 + 1 x + 1
j) x3 + 1
c) 49x4 − 1
g) x4 − 2x2 + 1
k)
d) x7 − 100x5
h) (x + 3)2 + 2(x + 3) + 1
l) 64x10 + x7
9
81
3
i) x3 − 27
4
1 x3 − 8
27
5. a)
Decompose the polynomial x3 + 5x2 + 3x + 15 into factors, and then justify that
it assumes positive values only for x > −5.
b) Decompose the polynomial 4x3 − 8x2 + 3x − 6 into factors, and then determine
for which values x this polynomial takes negative values.
c) Decompose the polynomial −12x5 + 6x4 − 2x + 1 into factors, and then justify that
for negative values of x the polynomial assumes positive values.
14
POLYNOMIALS
MLR2-1 str. 14
POLYNOMIAL EQUATIONS
Examples of
polynomial equations:
7x2 = 4x
−15x + 6 = 6x
(3x + 2)2 = 9x(x − 2)
−3x2 + 5x − 2 = 0
5x3 − 2x2 + 15x − 6 = 0
−6x2 (x2 − 3) = 2x2 + 5
Each of the equations in the box next to it
can be transformed so that on one side of
the equation there is a polynomial, and on
the other — number 0.
The equation that can be written in the form
W (x) = 0, where W (x) is a polynomial of degree n, is called a polynomial equation of
degree n.
The number, which is the solution to the
polynomial equation W (x) = 0, is called the
root of the polynomial W (x) or the root of
the equation.
EXERCISE A a) Find the first-degree polynomial equations in the box above
and calculate their roots.
b) Find the second-degree polynomial equations in the box above and calculate
their roots.
Methods of finding the roots of a first-degree polynomial equations have
already been discussed in elementary school. You also know how to solve
second-degree polynomial equations.
We will now show how to find the roots of some higher degree polynomials.
EXERCISE B
Give numbers that satisfy the equation:
a) (x + 1)(x − 6) = 0
b) (x − 2)4 = 0
c) 3x(x − 1)(2x − 4) = 0
It is quite easy to solve the polynomial equation
W (x) = 0, when the polynomial W (x) is presented in the product form. It is enough to take
advantage of the fact that the product of factors is equal to zero if any of the factors is
equal to zero.
a×b=0
⇑
⇓
a = 0 or b = 0
Note. When solving polynomial equations, decomposing a polynomial into factors
of the lowest degree is not necessary. It is enough to break down the polynomial
into factors whose roots we can calculate.
POLYNOMIAL EQUATIONS
MLR2-1 str. 15
15
Solve the equation.
EXAMPLE 1
a) x 5 − 6x 4 = 40x 3
We transform the equation into the form W (x) = 0.
x 5 − 6x 4 − 40x 3 = 0
We decompose the polynomial W (x) into factors.
x 3 (x 2 − 6x − 40) = 0
x3 = 0
or x 2 − 6x − 40 = 0
∆ = (−6)2 − 4 × 1 × (−40) = 196
√
∆ = 14
x =0
6 − 14
= −4
2
x1 =
x2 =
6 + 14
= 10
2
x = 0 or x = −4 or x = 10
b) x 3 − 6x 2 − 3x + 18 = 0
x 2 (x − 6) − 3 (x − 6) = 0
(x − 6) x 2 − 3 = 0
x − 6 = 0 or x 2 − 3 = 0
or x 2 = 3
√
√
x = 3 or x = − 3
x =6
x = 6 or x =
√
√
3 or x = − 3
c) 4x 6 = x 2
4x 6 − x 2 = 0
x 2 (4x 4 − 1) = 0
x 2 (2x 2 − 1)(2x 2 + 1) = 0
x2 = 0
or
2x 2 − 1 = 0
contradictory
equation
1
x =0
x2 = 2
x=
√
2
x = 0 or x = 2
PROBLEM
4
2x 2 +1 = 0
or
r
r
1
or x = −
2
√
2
or x = − 2
1
2
Solve the equation.
3
a) 2x − 5x − 3x2 = 0
16
b) 2x3 − 3x2 − 2x + 3 = 0
c) 15x3 = 9x
POLYNOMIALS
MLR2-1 str. 16
Solve the equation.
EXAMPLE 2
a) x 4 − 7x 2 + 12 = 0
For x 2 we substitute t and solve the equation
obtained with the unknown t.
x2 = t
t 2 − 7t + 12 = 0
√
∆=1
∆ = 72 − 4 × 12 = 1
t1 =
7−1
=3
2
2
t2 =
7+1
=4
2
We solve the equations x 2 = t1 and x 2 = t2 .
2
x =3
or
x =4
√
√
x = 3 or x = − 3 or x = 2 or x = −2
b) x 5 − 3x 4 − 8x 3 + 24x 2 − 9x + 27 = 0
x 4 (x − 3) − 8x 2 (x − 3) − 9(x − 3) = 0
(x − 3)(x 4 − 8x 2 − 9) = 0
x −3=0
x 4 − 8x 2 − 9 = 0
or
x2 = t
x =3
t 2 − 8t − 9 = 0
∆ = (−8)2 − 4 × (−9) = 100
t1 =
x 2 = −1
or
contradictory
equation
8 − 10
= −1
2
t2 =
√
∆ = 10
8 + 10
=9
2
x2 = 9
x = 3 or x = −3
x = 3 or x = −3
c)
2
x2 + 3 − 6 x2 + 3 − 7 = 0
x2 + 3 = t
t 2 − 6t − 7 = 0
∆ = (−6)2 − 4 × (−7) = 64
t1 =
x2 + 3 = 7
6+8
=7
2
t2 =
√
∆=8
6−8
= −1
2
x 2 + 3 = −1
or
x2 = 4
x 2 = −4
x = 2 or x = −2
contradictory equation
x = 2 or x = −2
PROBLEM
Solve the equation.
a) x4 − 9x2 + 20 = 0
POLYNOMIAL EQUATIONS
MLR2-1 str. 17
b) x5 + 2x4 + 3x3 + 6x2 − 4x − 8 = 0
c) (x2 − 7)2 + (x2 − 7) − 6 = 0
17
Let us now consider what the relationship between the degree of a polynomial and the number of roots of this polynomial may be.
It is known that the first-degree polynomial has one root (each equation of
the form ax + b = 0, where a 6= 0, has one solution). It is also known that a
second-degree polynomial can have two roots or one or no root at all.
EXERCISE C a) Each of the following three polynomials is a third-degree polynomial. Determine how many roots these polynomials have.
U(x) = x(x − 2)(x + 3)
V (x) = (x + 1)(x2 − 3x + 5)
W (x) = x(x + 5)2
b) Give an example of a fourth-degree polynomial that has no roots, and a
fifth-degree polynomial that has only one root.
Note that:
• A n-th degree polynomial has at most n roots (such a polynomial can be
broken down into at most n first-degree polynomials).
• Because each polynomial can be decomposed into factors of at most
second-degree, so in the case of an odd-degree polynomial, at least one
factor must be of first degree. Therefore, an odd-degree polynomial must
have at least one root.
Therefore, for example, a third-degree polynomial always has some root,
but it can’t have more than three. The fourth-degree polynomial may not
have roots, but if it has roots, it must be no more than four.
EXERCISE D Give an example of a polynomial of the lowest degree possible,
which has six roots.
Consider the following polynomials:
W (x) = (x − 7)(x − 5)2
P (x) = (x − 7)2 (x − 5)3
The roots of each of these polynomials are numbers 7 and 5.
In the decomposition of the polynomial W (x) into factors, the binomial x − 7 occurs once, and the binomial x − 5 occurs twice, because
W (x) = (x − 7)(x − 5)(x − 5). We say that the number 7 is a simple root of
the polynomial W (x), and the number 5 is its double root.
Note that in the decomposition of the polynomial P (x) into factors, binomial x − 7 occurs twice, and binomial x − 5 occurs three times. The number
7 is a root with multiplicity 2 (double root) of the polynomial P (x), and the
number 5 is a root with multiplicity 3 (threefold root) of this polynomial.
18
POLYNOMIALS
MLR2-1 str. 18
Let W (x) be a non-zero polynomial. Number a is called a root with
multiplicity k (short, k-fold root) of the polynomial W (x), when this
polynomial can be represented in the form:
W (x) = (x − a)k × P (x),
where P (x) is a polynomial and number a is not its root (P (a) 6= 0).
EXERCISE E
Number −1 is a threefold root of the polynomial W (x), and number 7 is its fivefold root. What can be said about the degree of polynomial
W (x)?
In the previous topic, we reminded formulas that allow you to calculate
roots of a quadratic polynomial ax2 +bx+c, where a 6= 0. Such a polynomial
can have two roots (and each of these roots is a simple root) or may have
one root (and this is a double root) or may have no roots at all. It is a
corollary of the following property: Polynomial W (x) = ax2 + bx + c , where
a 6= 0 , has two roots x1 and x2 if and only if W (x) = a(x − x1 )(x − x2 )
and x1 6= x2 , or one root x0 if and only if W (x) = a(x − x0 )2 .
EXERCISE F
Find the roots of the polynomial and determine their multiplicity.
a) x3 (x − 3)(x + 1)4
c) (x − 1) x2 − 6x + 5
b) (x + 2)2 (x + 5)3 (x + 2)
d) (x − 3) x2 − 6x + 9
PROBLEMS
1. Find all numbers satisfying the equation.
a) (x − 3)(2x + 5)(4 − 3x)2 = 0
b) (x + 5) x2 + x − 20 x2 − 5 = 0
c) x 2x2 + 9x + 9 9x2 + 1 = 0
d) 4x2 − 8x + 6 4x2 − 8x (−8x + 6) = 0
e) x3 x3 − 1 1 + x3 = 0
f) x2 + 2 x3 + 2 x3 + 8 = 0
2. Without solving the equation determine if it has solutions.
a) x4 + 1 = 0
POLYNOMIAL EQUATIONS
MLR2-1 str. 19
b) 3x2 + x4 = 0
c) 3x2 + 4x8 + 2 = 0
19
3. Solve the equation.
a) 5x3 − 45x = 0
c) x4 − 3x3 − 10x2 = 0
e) 2x5 − 14x4 + 24x3 = 0
b) 3x3 = x2 + 2x
d) 6x4 + 3x3 = 45x2
f) −5x4 + 3x3 + 14x2 = 0
4. Solve the equation.
a) 6x3 + 6x2 − 3x − 3 = 0
e) 2x5 − 8x3 + 16x2 − 64 = 0
b) 2x5 − 18x3 + 2x2 − 18 = 0
f) 3x5 − 12x3 − 12x2 + 48 = 0
c) 4x3 − 14x2 + 6x − 21 = 0
g) 5x5 + x3 − 6 = 30x2
d) 15x5 − 10x4 − 6x + 4 = 0
h) 5 = 3x + 5x4 − 3x5
5. Solve the equation.
a) x4 + x3 − 7x2 − x + 6 = 0
d) 3x4 + 6x3 + 4x2 + 2x + 1 = 0
b) x4 − 2x3 − 3x2 + 8x − 4 = 0
e) 2x4 + 2x3 + x2 − x − 1 = 0
c) x4 + 4x3 − 4x2 + 4x − 5 = 0
f) 2x4 + 6x3 + 3x2 − 3x − 2 = 0
6. Solve the equation.
a) x4 − 13x2 + 36 = 0
d) x5 + x3 − 12x = 0
g) x6 − 7x3 − 8 = 0
b) x4 − 9x2 + 20 = 0
e) 2x5 + 5x3 − 12x = 0
h) x7 + 5x4 + 6x = 0
c) 4x4 − 5x2 + 1 = 0
f) 3x5 − 8x3 − 3x = 0
i) 2x7 − x4 − x = 0
7. Find all the roots of the given polynomial and determine their multiplicity.
a) x7 (x − 1)3 (x + 2)(x + 5)5
c) (x + 2)4 (3x + 4)2 (x + 2)3
b) x(x + 3)2 (2x − 1)3 (x + 3)
d) (x − 3)2 (3 − x)3 (x + 3)2
8. Find all the roots of the given polynomial and determine their multiplicity.
2 2
x2 − 9 x2 + 2x − 15
x − 2x + 3
b) x2 + 6x + 9 2x2 + 9 2x2 + 5x − 3
c) (x − 1) x5 − 5x4 + 4x3
a)
d)
e)
f)
3x4 − x3 + 3x − 1 (x + 1)3
2
x2 − 1 x6 − 2x5 + x4
x3 − x2 x6 + x4 − x2 − 1
9. Number a is a k-fold
root of polynomial W (x) and a m-fold root of polynomial
V (x) (k > m). Justify that the number a is also a root of the Z(x) polynomial, and
determine its multiplicity if:
a) Z(x) = W (x) × V (x)
20
2
b) Z(x) = [W (x)] (V (x) + 3)
c) Z(x) = W (x) + V (x)
POLYNOMIALS
MLR2-1 str. 20
DIVISION OF POLYNOMIALS
It is known that if a given natural number a is the product of some two
numbers, as a result of dividing number a by one of these numbers, we
get the second one. For example, the equality:
4503 = 57 × 79
we can express as:
4503 ÷ 57 = 79
This means that the number 4503 is divisible by 57 (and also by 79). We
will understand polynomials’ division in a similar way.
You already know that polynomials can be broken down into factors. For
example, the polynomial W (x) = 2x3 −4x2 +3x−6 can be written as a product:
2x3 − 4x2 + 3x − 6 = (x − 2)(2x2 + 3)
The above equality can be written differently:
(2x3 − 4x2 + 3x − 6) ÷ (x − 2) = 2x2 + 3
We say then that the polynomial W (x) is divisible by the polynomial x − 2.
The result of dividing the W (x) polynomial by x − 2 is 2x2 + 3 polynomial.
Note. The polynomial W (x) is also divisible by the 2x2 + 3 polynomial.
We say that the polynomial W (x) is divisible by the non-zero polynomial P (x), if there is such a Q(x) polynomial that:
W (x) = P (x) × Q(x)
This equality can also be written in the form:
W (x) ÷ P (x) = Q(x)
Note that after decomposing a given polynomial into factors, it is easy to
indicate polynomials by which it is divisible and give the results of such
division.
EXERCISE A Decompose the polynomial x5 − 4x3 + x2 − 4 into factors, and then
determine the division result.
a) (x5 − 4x3 + x2 − 4) ÷ (x − 2)
b) (x5 − 4x3 + x2 − 4) ÷ (x2 − 4)
We will now show a method that allows you to find the result of dividing two polynomials without having to break down the first of them into
factors. This method resembles the division algorithm of natural numbers.
DIVISION OF POLYNOMIALS
MLR2-1 str. 21
21
The result of the division W (x) ÷ P (x), where W (x) = 2x3 − x2 − 16x + 3
and P (x) = x − 3, can be found as follows:
EXERCISE B Make sure the result is correct — multiply the polynomial x − 3 by
the polynomial 2x2 + 5x − 1.
22
POLYNOMIALS
MLR2-1 str. 22
Note that when performing the division, we calculated the differences of
some polynomials. When we do such operations, it is easy to make a mistake, so it is worth slightly change the way of writing.
Divide the polynomial W (x) = 3x 3 + 16x 2 + 3x − 10 by the binomial
x + 5 and write the polynomial W (x) as a product.
EXAMPLE 1
3x 2 + x − 2
(3x 3 + 16x 2 + 3x − 10) ÷ (x + 5)
−3x 3 − 15x 2
x 2 + 3x − 10
−x 2 − 5x
−2x − 10
2x + 10
0
W (x) = (x + 5)(3x 2 + x − 2)
Perform the division (5x3 − 17x2 + 7x − 3) ÷ (x − 3).
PROBLEM
Similarly to dividing natural numbers, also by dividing a polynomial by
another polynomial we can get a remainder different from 0.
Numbers
211
2745 ÷13
− 26
14
− 13
15
− 13
2 ←− remainder
Polynomials
2
2x + x + 1
(2x3 + 7x2 + 4x + 5) ÷ (x + 3)
−2x3 − 6x2
x2 + 4x + 5
−x2 − 3x
x+5
−x − 3
2
←− remainder
Result of division:
Result of division:
2745 ÷ 13 = 211 remainder 2
(2x3 + 7x2 + 4x + 5) ÷ (x + 3) = 2x2 + x + 1 remainder 2
So:
So:
2745 = 211 × 13 + 2
The remainder is smaller from
the number we divide by.
DIVISION OF POLYNOMIALS
MLR2-1 str. 23
3
2
2x + 7x + 4x + 5 = (x + 3)(2x2 + x + 1) + 2
The degree of the remainder is smaller than
the degree of the polynomial we divide by.
23
Perform the division W (x) ÷ P(x), where W (x) = x 5 − 4x 4 − 3x 2 +
+14x − 3, P(x) = x − 4.
EXAMPLE 2
x 4 − 3x + 2
(x 5 − 4x 4 − 3x 2 + 14x − 3) ÷ (x − 4)
−x 5 + 4x 4
−3x 2 + 14x − 3
3x 2 − 12x
2x − 3
−2x + 8
5
As the result of division
W (x) ÷ P(x) we receive x 4 − 3x + 2
remainder 5.
(x 5 − 4x 4 − 3x 2 + 14x − 3) ÷ (x − 4) = x 4 − 3x + 2 remainder 5
So:
x 5 − 4x 4 − 3x 2 + 14x − 3 = (x − 4)(x 4 − 3x + 2) + 5
PROBLEM
W (x) = P(x) × (x 4 − 3x + 2) + 5
Perform the division (2x3 + x2 − 12x + 5) ÷ (x + 3).
Note that the remainder of dividing any W (x) polynomial by the binomial
x − a is always a number. When this remainder is 0, we say that W (x) is
divisible by the binomial x − a.
PROBLEMS
1. Give
examples of three polynomials by which both polynomials W (x) and V (x)
are divisible.
a) W (x) = −(3x + 2)4 (2x + 5)3 , V (x) = (2x + 5)3 (x + 2)
b) W (x) = 1 (2x − 1)5 (4x − 1)(x2 + 2), V (x) = (x2 + 2)5 (2x − 1)2 (x − 4)
3
2. Perform the division.
a)
b)
c)
d)
e)
f)
x3 − 8x2 + 17x − 10 ÷ (x − 5)
3x3 + 8x2 − 18x − 8 ÷ (x + 4)
x4 − 2x3 + x2 + x − 1 ÷ (x − 1)
−4x4 + 12x3 − 5x2 + 17x − 6 ÷ (x − 3)
x5 + 9x4 + 14x3 + x + 7 ÷ (x + 7)
−x5 − 2x4 + 5x2 + 17x + 14 ÷ (x + 2)
24
POLYNOMIALS
MLR2-1 str. 24
3. Find a polynomial W (x) such that there is the given equality.
a) (x + 2) × W (x) = 2x3 + 9x2 + 7x − 6
b) W (x) × (x − 3) = 5x4 − 17x3 + 6x2 − x + 3
c) 10x2 + (x − 5) × W (x) = 3x4 − 13x3 − x + 5
d) 7(2x2 + x) − (x + 4) × W (x) = 4 − 3x3
4. What numbers should be inserted in place of the boxes?
a)
3x3 +
x4 +
b)
x2 +
x3 +
÷ (x + 2) = x2 + 1
x2 − 15x ÷ (x − 3) = −2x3 +
x+
x
5. Perform the division with remainder.
a)
b)
c)
d)
e)
f)
x3 − 4x2 + 10x ÷ (x − 1)
2x3 + 7x2 + 2x − 1 ÷ (x + 3)
3x4 + 4x3 − 9x2 − 11x − 4 ÷ (x + 2)
−4x4 + 17x3 − 4x2 + 2x − 5 ÷ (x − 4)
x5 + x4 − 2x2 + x + 10 ÷ (x + 1)
3x5 − 13x4 − 10x3 − x2 + 7x − 6 ÷ (x − 5)
6. Find the polynomial W (x) satisfying the given condition.
a) By dividing W (x) by x − 4, we get the 3x − 2 polynomial and the remainder −2.
b) By dividing W (x) by x + 5, we get the polynomial x2 − 5 and the remainder 7.
7. Give an example of a fourth-degree polynomial for which the remainder of its
division by the polynomial x + 2 is equal to 10.
8. a) The remainder of the division of the polynomial W (x) = 2x3 − x2 − 11x − 4 by
the binomial P (x) = x − 2 is equal to −14. What number should the free term of the
polynomial W (x) be replaced by so that the obtained polynomial be divisible by the
P (x) binomial?
b) The remainder of the division of the polynomial V (x) = 5x4 + 15x3 − 2x2 − 2x + 7
by the binomial Q(x) = x + 3 is −5. How should the free term of the V (x) polynomial
be changed so that the obtained polynomial be divisible by the Q(x) binomial?
9. a)
Given are polynomials:
W1 (x) = 2x3 − 11x2 − 4x − 2
W2 (x) = 3x4 − 15x3 + 4x − 16
Divide each of these polynomials by x − 5 binomial, then divide the W1 (x) + W2 (x)
polynomial by the x − 5 binomial.
b) Prove that if the remainder of division of the polynomial W (x) by x − a is r1 and
the remainder of division of the polynomial V (x) by x − a is r2 , then the remainder
of division of the polynomial W (x) + V (x) by x − a is equal r1 + r2 .
DIVISION OF POLYNOMIALS
MLR2-1 str. 25
25
Horner’s scheme
By dividing a fourth-degree polynomial by the x − p binomial, we get a third-degree
polynomial and the remainder r (the remainder can be 0).
So we can write the equality:
a4 x4 + a3 x3 + a2 x2 + a1 x + a0 = (x − p)(b3 x3 + b2 x2 + b1 x + b0 ) + r
Transforming the right-hand side of this equality, we get:
b3 x4 + (b2 − pb3 )x3 + (b1 − pb2 )x2 + (b0 − pb1 )x + r − pb0
From equality of polynomials it follows that:
a4 = b3
a3 = b2 − pb3
a2 = b1 − pb2
a1 = b0 − pb1
a0 = r − pb0
b2 = pb3 + a3
b1 = pb2 + a2
b0 = pb1 + a1
r = pb0 + a0
Hence:
b3 = a4
Similar relationships would be obtained by dividing any polynomial by the binomial
x − p. Therefore, looking for the result of such a division, we can use a simplified
method called the Horner’s scheme.
It is shown below by dividing the −x4 −4x3 +2x2 −25 polynomial by the x+4 binomial.
William Horner was a British mathematician who popularized this method.
By dividing the polynomial −x4 − 4x3 + 2x2 − 25 by the x + 4 binomial we get the
polynomial −x3 + 2x − 8 and the remainder 7, i.e.:
(−x4 − 4x3 + 2x2 − 25) = (x + 4)(−x3 + 2x − 8) + 7
10. Write down any fifth-degree polynomial. Then divide it by the x + 2 binomial.
Perform this division in two ways: using the method we learned earlier and the
Horner’s scheme shown above.
26
POLYNOMIALS
MLR2-1 str. 26
11. Perform the division using Horner’s scheme.
a)
b)
c)
x3 − 6x2 + 12x − 8 ÷ (x − 2)
x4 + x3 + x2 + 4x + 3 ÷ (x + 1)
2x4 − 8x3 + 5x − 20 ÷ (x − 4)
d)
e)
f)
x5 − 9x3 + 2x + 5 ÷ (x + 3)
5x5 − 7x4 − x3 + 4x2 + 3x ÷ (x − 1)
−3x4 + 2x − 3 ÷ (x + 2)
REMAINDER THEOREM
We already know that by dividing any polynomial W (x) by the binomial
x − a, we get a certain polynomial Q(x) and the remainder R, which is a
number. The polynomial W (x) can be written as:
W (x) = (x − a) × Q(x) + R
When R = 0, then the equality occurs:
W (x) = (x − a) × Q(x)
Then W (a) = 0, i.e. a is the root of the polynomial W (x).
The general property of polynomials that connects the root of a polynomial to the division of this polynomial by a certain binomial is expressed by
the Remainder Theorem.
Remainder Theorem
Number a is the root of a polynomial W (x)
if and only if when
this polynomial is divisible by the x − a binomial.
Note. The above theorem can be written as follows:
W (a) = 0
⇐
⇒ W (x) ÷ (x − a) = P (x) ,
where P (x) ) is a certain polynomial.
Proof
Note that for any number a polynomial W (x) can be written in the form:
W (x) = (x − a) × P (x) + R,
where P (x) is a certain polynomial and R — a certain number.
Therefore:
W (a) = (a − a) × P (a) + R = 0 × P (a) + R = R
We have therefore obtained the equality: W (a) = R
REMAINDER THEOREM
MLR2-1 str. 27
27
By using this equality, we will prove two implications.
1. Suppose the number a is a root of the polynomial W (x), i.e. W (a) = 0.
Because W (a) = R, therefore R = 0. It follows that the polynomial W (x) is
divisible by the binomial x − a.
2. Suppose now that the polynomial W (x) is divisible by x − a, i.e. R = 0.
Because W (a) = R, therefore W (a) = 0. It follows that a is a root of polynomial
W (x).
EXERCISE A A polynomial W (x) = x4 + x3 − 7x2 − x + 6 is given. Using the Remainder Theorem, determine by which of the given binomials it is divisible.
x−1
x+1
x−2
x+2
Analyzing the proof of Remainder Theorem, it can be seen that, by the
way, the following property of polynomials has been justified:
The remainder of division
of polynomial W (x) by x − a binomial
is equal to W (a).
EXERCISE B
by x − 2.
Determine the remainders from dividing the given polynomials
A(x) = x3 − x2 + 3x − 5
B(x) = 1 x5 − x2 − 7x + 1
4
C(x) = x4 − 6x3 + 5x2 + 12
Suppose we are to solve the polynomial equation W (x) = 0 and we know
that number p is a root of the polynomial W (x).
The Remainder Theorem implies that the polynomial W (x) is divisible by
x − p. So, there exists a polynomial Q(x) of degree lower than W (x), such
that:
W (x) = (x − p) × Q(x)
The equation W (x) = 0 can be written as:
(x − p) × Q(x) = 0
To determine if the W (x) polynomial has any other roots, you can solve
the lower-degree equation: Q(x) = 0.
28
POLYNOMIALS
MLR2-1 str. 28
Check that number 1 is a solution to the given equation, and find
other solutions to this equation.
EXAMPLE
x 3 − 3x 2 + 2 = 0
13 − 3 × 12 + 2 = 1 − 3 + 2 = 0
We check if number 1 meets the equation
x 3 − 3x 2 + 2 = 0.
x 2 − 2x − 2
(x 3 − 3x 2 + 2) ÷ (x − 1)
−x 3 + x 2
−2x 2 + 2
2x 2 − 2x
−2x + 2
2x − 2
We divide the polynomial x 3 − 3x 2 + 2 by the
binomial x − 1, to decompose the polynomial
into factors.
0
3
2
x − 3x + 2 = (x − 1)(x 2 − 2x − 2)
(x − 1)(x 2 − 2x − 2) = 0
x =1
or
We write the equation in a different form and
solve it.
x 2 − 2x − 2 = 0
∆ = 4 − 4 × (−2) = 12
√
√
√
∆ = 12 = 2 3
√
√
2−2 3
=1− 3
2
√
√
2+2 3
x2 =
=1+ 3
2
x1 =
Ans. The equation has three solutions: x1 = 1 −
√
3, x2 = 1 +
√
3, x3 = 1.
PROBLEM
Check that 2 is a root of the polynomial W (x) = x3 − 7x + 6, and then
solve the equation W (x) = 0.
Curiosity
In Poland, the Remainder Theorem is called Bézout’s theorem. Étienne
Bézout (1730 –1783) was a French mathematician. He did algebra, but he
is known mainly as the author of excellent textbooks, praised as exceptionally clear. Bézout’s Cours de mathématiques, translated into English, was
for many years a basic textbook at Harvard University. The Remainder Theorem was neither formulated nor proved by Bézout — it was already known
by then.
REMAINDER THEOREM
MLR2-1 str. 29
29
PROBLEMS
1. Is the W (x) polynomial divisible by the V (x) binomial?
a) W (x) = 5x14 − 6x + 1,
V (x) = x − 1
b) W (x) = 3x7 − x3 + x2 + 1,
V (x) = x + 1
c) W (x) = x3 + 3x2 + x − 10,
V (x) = x + 2
d) W (x) = x4 − 1 x3 − 4x2 + 1,
V (x) = x − 1
2
2
2. Without performing the division determine what is the remainder of division of
the polynomial W (x) by the binomial P (x).
a) W (x) = 3x5 − 2x3 + 4
P (x) = x − 1
b) W (x) = −5x3 − 3x2 + 6
P (x) = x + 2
3. a)
For what value of a the polynomial 5x5 − ax3 + 3x2 − 6x is divisible by the
binomial x − 2?
b) For what value of p the polynomial px5 − px3 − 1 x + 2 is divisible by the binomial
2
x + 2?
4. Check that the given number is a root of the equation, and then find its remaining roots.
a) 2x3 − x2 − 8x + 4 = 0,
b) 6x3 − 29x2 − 6x + 5 = 0,
c) x3 + 7x2 − 5x − 75 = 0,
d) 4x3 − 4x2 − 15x + 18 = 0,
2
5
3
−2
e) x4 − x3 − 14x2 + 2x + 24 = 0,
f) x4 + 8x3 + 19x2 + 32x + 60 = 0,
POLYNOMIAL EQUATIONS
−3
−5
(CONTINUED)
To use the Remainder theorem in solving a polynomial equation, you must
know at least one number that satisfies this equation. Even when such a
number exists, it is generally difficult to find.
We will now try to discover what conditions would have to be met so as
one of the solutions of the polynomial equation with integer coefficients
be a rational number.
30
POLYNOMIALS
MLR2-1 str. 30
Let’s consider, for example, a polynomial:
W (x) = 5x3 − 6x2 + 16x − 3
The roots of this polynomial satisfy the equation:
5x3 − 6x2 + 16x − 3 = 0
, where p and q are integers, is a solution
Suppose the rational number p
q
to this equation. We can assume that the fraction p
is simplified, that is,
q
the numbers p and q do not have common divisors different from 1 and
−1 (they are relatively prime). So, there is equality:
2
3
5 × p3 − 6 × p2 + 16 × p − 3 = 0
q
q
q
We can write this equality in the form:
5p3 − 6p2 q + 16pq 2 − 3q 3 = 0
We will transform the obtained equality in two ways:
−3q 3 = −5p3 + 6p2 q − 16pq 2
5p3 = 6p2 q − 16pq 2 + 3q 3
Hence:
−3q 3 = p −5p2 + 6pq − 16q 2
5p3 = q 6p2 − 16pq + 3q 2
The numbers p and q are integers, so on both sides of each of the two
equations obtained is the product of integers. The first equality shows that
the number −3q 3 is divisible by p. The second equality shows that the
number 5p3 is divisible by q.
We assumed that the numbers p and q are relatively prime, so:
Number p is a divisor of −3, which is the free term of W (x) polynomial, i.e.
p ∈ {1, −1, 3, −3}.
Number q is a divisor of 5, which is the coefficient at the highest power of
the polynomial W (x), i.e. q ∈ {1, −1, 5, −5}.
So, if the polynomial under consideration has a rational root
of the numbers:
1, −1, 3, −3, 1 , − 1 , 3 , − 3
5
5
5
p
,
q
this is one
5
It’s easy to check that none of the numbers 1, −1, 3, −3 is a root of this
1
polynomial, and number 5 is its root.
3
2
W 1 = 5 × 1 − 6 × 1 + 16 × 1 − 3 = 0
5
5
5
5
By checking the value of a polynomial for the remaining fractions, all
rational roots of this polynomial can be found.
POLYNOMIAL EQUATIONS
(CONTINUED)
MLR2-1 str. 31
31
Theorem (on rational roots)
Suppose in a polynomial equation:
an xn + an−1 xn−1 + . . . + a1 x + a0 = 0
all coefficients are integers, a0 6= 0 and an 6= 0. If a solution to this
equation is a rational number, then it can be presented in the form
p
of a fraction q , where the numerator p is a divisor of the free term
a0 , and the denominator q is the divisor of the coefficient an at the
highest power.
Proof
Let’s assume that the numbers an , an−1 , . . . , a2 , a1 , a0 in the following equation
are integers and a0 6= 0 and an 6= 0.
an xn + an−1 xn−1 + . . . + a2 x2 + a1 x + a0 = 0
p
Suppose a certain number q , where p and q are integers other than 0, is a
p
solution to the equation. We can also assume that the fraction q is simplified
(that is, p and q do not have common divisors different from 1 and from −1).
So the equality is met:
n
an p n + . . . + a1 p + a0 = 0
q
q
We will first show that p is a divisor of a0 .
After multiplying both sides of the equality by q n we get:
an pn + an−1 pn−1 q + . . . + a1 p1 q n−1 + a0 q n = 0
Let’s write this equality in the form:
p(an pn−1 + an−1 pn−2 q + . . . + a1 q n−1 ) = −a0 q n
We have assumed that all numbers labeled with letters in this equality are
integers, so on both sides is the product of integers. Number q n has no common divisors with number p (different from 1 and from −1). It follows that
number a0 must be divisible by number p. We have therefore shown that p is
the divisor of the free term a0 .
Let us now write the equality obtained earlier in the form:
q(an−1 pn−1 + . . . + a1 pq n−2 + a0 q n−1 ) = −an pn
Reasoning as above, we will come to the conclusion that q is a divisor of the
an coefficient.
EXERCISE A Number
5
2
5
is a solution to one of the equations given. To which one?
3
6x + 3x − 2x − 4 = 0
3
2
10x + 7x − 5 = 0
32
5x3 + 8x2 − 14x + 4 = 0
x4 + 5x2 − x + 10 = 0
POLYNOMIALS
MLR2-1 str. 32
Note that if the polynomial equation with integer coefficients has no rational solutions, it doesn’t mean that there are no solutions at all.
EXERCISE B Give an example of such a polynomial equation with integer coefficients that has a solution but no rational solutions.
Note that when looking for integer roots of a polynomial, we can use the
following conclusion from the above theorem.
Theorem (on integer roots)
Suppose in a polynomial equation:
an xn + an−1 xn−1 + . . . + a1 x + a0 = 0
all coefficients an , an−1 , . . . , a0 are integers and a0 6= 0. If the solution
to this equation is an integer, it is the divisor of the free term a0 of the
polynomial.
Curiosity
At the beginning of the year 1535 a mathematical duel began. Its participants were two Italians — Mario Fior and Niccolò Fontana, known by the
nickname Tartaglia. Such scientific competitions were very popular then,
because thanks to them the winner’s fame (and income) grew.
The competition consisted in that within 30 days both participants were to
solve dozens of problems prepared by the opponent. Fior had already won
many tournaments before, because he knew several third-degree equations
that only he seemed to be able to solve. All Fior’s problems concerned
third degree equations. A week before the deadline, Tartaglia discovered
a method for solving this type of equation and won the tournament.
Many scholars wanted to know the Tartaglia’s method, but he did not
want to reveal it to anyone. One of the Italian mathematicians — Girolamo Cardano — asked Tartaglia for so long to show his method that he
finally agreed — but on condition that Cardano will not disclose it to anyone. A few years later, Cardano learned that the Tartaglia’s method had
been previously discovered by Scipione del Ferro. He decided then that his
promise to Tartaglia was no longer valid for him, and he dishonestly published the method under his own name. Since then, the formula for solving
the cubic equation is called Cardano’s formula.
When solving a polynomial equation with integer coefficients, it is worth
to start by checking if there are integer solutions. This way you can often
simplify your work.
POLYNOMIAL EQUATIONS
(CONTINUED)
MLR2-1 str. 33
33
Solve the equation: 2x 3 − 5x 2 + x + 3 = 0.
EXAMPLE
The divisors of the free term of W (x) = 2x 3 − 5x 2 + x + 3 are the numbers 1, −1,
3, −3.
W (1) = 2 − 5 + 1 + 3 6= 0
W (−1) = 2 × (−1) − 5 − 1 + 3 6= 0
We check if the equation has integer roots.
W (3) = 2 × 27 − 5 × 9 + 3 + 3 6= 0
W (−3) = 2 × (−27) − 5 × 9 − 3 + 3 6= 0
p
The devisors of the coefficient
at the highest power are:
If the equation has a rational solution q , then
the numerator is a divisor of 3 and the denominator is a divisor of 2; we list all the
p
numbers q such that p ∈ {1, −1, 3, −3} and
q ∈ {1, −1, 2, −2}.
1, −1, 2, −2
Possible rational solutions:
1
1
3
3
1, −1, 2 , − 2 , 2 , − 2 , 3, −3
1
2
W
W
1
1
1
= 2 × 8 − 5 × 4 + 2 + 3 6= 0
1
1
1
1
8
W − 2 = 2 × − 8 − 5 × 4 − 2 + 3 = − 4 + 3 6= 0
3
2
3
2
3
3
3
27
45
18
= 2× 2 − 5× 2 + 2 + 3 = 4 − 4 + 4 = 0
We have already checked
that there is no integer
solution of the equations,
therefore, it is enough to
check if among the numbers 12 , − 12 , 32 , − 32 is a
solution.
2x 2 − 2x − 2
3
(2x 3 − 5x 2 + x + 3) ÷ x −
2
3
2
−2x + 3x
We divide the polynomial by the binomial
x − 23 , to factorize the polynomial.
−2x 2 + x + 3
2x 2 − 3x
−2x + 3
2x − 3
0
We write the equation in a different form,
and then solve it.
3
x − 2 (2x 2 − 2x − 2) = 0
x−2 =0
3
or 2x 2 − 2x − 2 = 0
3
x 2 − x − 1 = 0,
x= 2
x1 =
√
1− 5
,
2
x2 =
÷2
∆=5
√
1+ 5
2
Ans. The equation has three solutions: x1 =
PROBLEM
√
√
1− 5
1+ 5
3
,
x
=
, x3 = 2 .
2
2
2
Solve the equation: x3 + 4x2 − 11x − 30 = 0.
34
POLYNOMIALS
MLR2-1 str. 34
The method of solving equations described in this chapter can also be used
to solve equations with rational (not necessarily integer) coefficients. It is
enough to note that when we multiply both sides of the equation by the
common denominator of all coefficients, we get the equivalent equation
with integer coefficients.
EXERCISE C
equation:
Write the equation with integer coefficients equivalent to the
1 4
x
3
− 13 x3 + 21 x2 − 61 x +
1
6
=0
Justify that this equation has no integer roots.
PROBLEMS
1. One of the five numbers given is the root of the polynomial W (x). Which?
a) W (x) = 5x3 + 23x2 − 35x + 10
1,
3
b) W (x) = 3x5 − x4 − 6x3 + 2x2 − 45x + 15
−1,
−3,
5
2
1,
3
c) W (x) = −6x − 11x + 13x + 15
3,
5
−5,
6
d) W (x) = −2x5 − 9x4 − 9x3 − 11x2 − 10x + 14
−3,
3
2
7
1 ,
14
2,
5
−4,
3
5
−1,
5,
2
4
2,
− 15
8
6
5
−4,
−3 1 ,
2
2,
7
−12
2. Find all integer roots of the polynomial.
a) x3 − 2x2 − 2x − 3
d) 2x3 + 8x2 + 8x + 6
b) 3x4 + x3 − x2 − x − 2
e) −x5 − 3x4 + 6x2 + 4x
c) x3 − 3x2 − 6x + 8
f) x6 − 4x5 − 6x4 + 4x3 + 5x2
3. For which natural values of n the equation xn + x + 2 = 0 has integer solutions?
Find these solutions.
4. Justify that equation
97x10 − x + 1 = 0 has no rational solutions.
5. For what integer values of m the polynomial
9x3 − mx + 1 has a rational root?
6. Among the rational roots of the polynomial
W (x) = 5x4 − 11x3 + ax2 + bx − 2 there
are two numbers that are roots of the polynomial Q(x) = 2x4 + cx3 + dx2 + 9x + 5.
Find the coefficients a, b, c and d.
POLYNOMIAL EQUATIONS
(CONTINUED)
MLR2-1 str. 35
35
Good to know!
The theorem on integer roots of
a polynomial with integer coefficients can be used to justify
the irrationality of some numbers. For example, we will show
the following theorem:
√
The number 2 is irrational.
Proof
Consider the equation x2 − 2 = 0.
The only ”candidates” for rational solutions to
this equation are numbers 11 , − 11 , − 21 and 21 . Since
none of them satisfies the equation x2 − 2 = 0, so
it has no rational solutions.
It is known that one of the solutions to this equa√
√
tion is the number 2. It follows that 2 is not a
rational number.
7. Read the proof in the box. Show that the numbers are irrational:
a)
√
5
√
b) −2 3
c)
√
3
7
d)
√
100
12
8. Show that if p is a prime number, then for the natural number n 6= 2 the number
√
n
p is irrational.
36
POLYNOMIALS
MLR2-1 str. 36
Figures
on the plane. Part 1
Geometry in Greek means „measuring the Earth”.
This domain of science originates from measuring farmlands.
But for thousands of years already people are occupied
with geometry not for its application but for pure curiosity.
Angles. Angles in triangles and quadrilaterals
of triangles
Pythagoras' theorem and inverse of Pythagoras' theorem
Properties of triangles (continued)
MLR2-1 str. 37
Basic properties
Properties of quadrilaterals
Some introductory remarks
Geometric figures are abstract objects, so when we say, for example, draw a straight line,
draw a triangle, we mean drawing a model of the figure being considered.
Points are labelled with capital letters. We can name lines with lowercase letters or with a
pair of uppercase letters. We usually denote rays with a pair of capital letters, where the
first of them is the point where the ray starts (confusingly called the endpoint).
Line segments can be labelled with lower
case letters. We assume that a lowercase
letter can also mean the length of a segment.
Line segments are also labelled with two
capital letters, e.g. AB, but the length of
the line segment named in this way is written as: |AB|.
With the symbol k we write that lines or
line segments are parallel, and with the
symbol ⊥ we write that lines or line segments are perpendicular.
|AB| = 2,4 cm
akb
AB ⊥ CD
We call polygons by several capital letters.
Letters mean subsequent vertices of the
polygon.
triangle ABC
pentagon KLMNO
When we say that the length of the segment is 4, it means that the segment is 4 times
longer than a certain segment taken as a unit of length. When we say that the area of the
figure is equal to 12, it means that the area of this figure is 12 times larger than the area
of a certain square taken as a unit of area.
Angles are most often labelled with lower-case letters of the Greek alphabet:
α (alpha)
β (beta)
γ (gamma)
δ (delta)
ϕ (phi)
χ (chi)
ψ (psi)
ω (omega)
We assume that a Greek letter can also mean the angle’s measure.
We can also name angles with three capital letters, of which the middle letter
always indicates the vertex of the angle.
E.g. ¾ AOB is an angle with the vertex
O. The measure of the angle ¾ AOB is
|¾ AOB|.
38
α = 24◦
¾ AOB
|¾ AOB| = 24◦
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 38
ANGLES. ANGLES IN TRIANGLES
AND QUADRILATERALS
Let’s recall that two rays with a common
endpoint divide the plane into two angles;
the rays are the arms of both angles. If
these arms are not on one line, then one
of the angles is a convex angle and the
other is a non-convex one. The sum of
their measures is 360◦.
β = 360◦ − α
D
E
The angle is convex when its measure belongs to the set 0◦ ; 180◦ ∪ {360◦}.
The right angle
has 90◦.
The α angle
is acute,
when α ∈ 0◦ ; 90◦ .
The straight angle
has 180◦.
The complete angle
has 360◦.
The α angle
is obtuse,
The α angle is non-convex,
when α ∈ 90◦ ; 180◦ .
when α ∈ 180◦ ; 360◦ .
EXERCISE A How many non-convex angles can
be indicated in the figure beside?
Good to know!
Figure F is called non-convex (or concave) when there is a line segment
with ends belonging to figure F , that is not included in this figure. If each
of the line segments with the ends belonging to figure F is contained in
this figure, then this figure is called convex.
Examples of non-convex figures
ANGLES. ANGLES IN TRIANGLES AND QUADRILATERALS
MLR2-1 str. 39
Examples of convex figures
39
EXERCISE B What measures have angles, which are labelled with letters ϕ,
χ and ψ in the picture?
Supplementary angles
γ = 180◦ − δ
Vertical angles
Angles that have a common arm and
together form the straight angle are
called supplementary angles. The sum
of supplementary angles’ measure is
180◦.
When the arms of one angle are extensions of the arms of the other angle,
such angles have equal measures and
we call them vertical angles. Two intersecting lines form two pairs of vertical angles.
EXERCISE C Draw two intersecting lines and indicate in this drawing two pairs
of vertical angles and four pairs of supplementary angles.
The line that intersects two parallel
lines is inclined to each of them at the
same angle.
The opposite is also true: if a line
crosses two other lines and each of
them at the same angle, the two lines
must be parallel.
EXERCISE D The lines m and n in the drawing are parallel. What measures have angles
labelled with letters?
If two lines are crossed by a third line, we can indicate four pairs of angles,
which we call corresponding, and four pairs of angles called alternate.
40
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 40
In each of the drawings below, a line intersects two parallel lines, so the
two marked angles have equal measures.
Corresponding angles
Alternate angles
We can use the recalled knowledge to prove the theorem about the angles
of a triangle.
Theorem
The sum of the angle measures
of a triangle is 180◦.
Proof
Let α, β and γ stand for the angles of triangle KLM. We will show that
α + β + γ = 180◦.
Through the vertex M of the KLM triangle
we draw a line parallel to the line KL. Let’s
denote by δ and ϕ the angles that together
with the angle γ form a straight angle, i.e.
δ + ϕ + γ = 180◦
The line KM intersects two parallel lines, so α = δ, because these are alternate
angles. Similarly, β = ϕ.
Thus: α + β + γ = δ + ϕ + γ = 180◦.
EXERCISE E
In an isosceles triangle one of the angles has 70◦. What measures
the other angles of the triangle have?
ANGLES. ANGLES IN TRIANGLES AND QUADRILATERALS
MLR2-1 str. 41
41
Because any quadrilateral can be divided
by one of the diagonals into two triangles, so the sum of the angle measures
of both these triangles is the sum of the
quadrilateral’s angles measures. The following statement is therefore true:
Theorem
The sum of the quadrilateral’s
angles measures is 360◦.
By using the properties of the corresponding or alternate angles, the following properties of the trapezoid and the parallelogram can be justified.
Theorem
In the trapezoid, the sum
of measures of the angles lying
at the same arm is 180◦.
EXERCISE F
Write a proof of the trapezoid’s angles measures theorem, modeled
on the proof of the triangle’s angles theorem. Use the picture beside
Theorem
In the parallelogram, the opposite angles
have the same measures, and the sum
of measures of the angles lying
at the same side is equal to 180◦.
α + β = 180◦
EXERCISE G In the drawing, the lines a
and b as well as m and n contain the
sides of a parallelogram, i.e. a k b and
m k n. Identify angles that have the same
measure as angle α, and angles that have
the same measure as angle β. Using this
drawing, prove the above theorem.
42
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 42
PROBLEMS
1. Using the information under the figure, calculate the angle measures α i β.
2. The T , O, P
points in the picture lie on one line
and |¾ SOR| = 50◦, |¾ T OR| = 110◦. Calculate the
non-convex angle’s P OS measure.
3. Look at the picture next to. Justify that
points A, B, C do not lie on one line.
4. The lines marked in blue are parallel. Calculate the angles labelled with letters.
5. What measures have the angles of the triangle marked in blue?
6. The
sun’s rays fall at an angle of
50◦ to the earth’s surface. The vertical stick casts a shadow. At what angle should the stick be tilted towards
the shadow so that the stick and its
shadow have the same length?
ANGLES. ANGLES IN TRIANGLES AND QUADRILATERALS
MLR2-1 str. 43
43
7. Look at the picture beside. The CD line segment
divides the isosceles triangle ABC (|AB| = |AC|) into
two isosceles triangles (|CD| = |AD| = |CB|). What is
the angle measure of α?
8. Points
A, B and C are not collinear. The measure of the angle adjacent to the
ABC angle is twice as large as the BAC angle measure. Justify that the triangle
ABC is an isosceles triangle.
9. What measures have the angles of the quadrilateral drawn?
10. What measures have angles in the drawn parallelogram?
11. What measures have angles in the drawn trapezoid?
12. The shorter diagonal divides a trapezoid into an equilateral triangle and a
right-angled triangle. Calculate the measures of the angles of this trapezoid.
13. Draw any rhombus and connect with line segments the centers of its adjacent
sides. Show that in the quadrilateral obtained in this way all angles are right (that
is, the quadrilateral is a rectangle).
44
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 44
BASIC PROPERTIES OF TRIANGLES
EXERCISE A a) In the triangle ABC |AB| = 12, |BC| = 14, |AC| = 22. Which
angle of this triangle is the largest?
b) In the KLM triangle data are | ¾ KLM| = 71◦ and | ¾ LKM| = 66◦. Which side
of this triangle is the longest?
In the previous chapter, we talked about the sum of measures of triangle’s
angles. It is also worth remembering that in the triangle the longest side
is opposite the largest angle, and the shortest side is opposite the smallest
angle.
Now we will recall two important properties of triangles.
EXERCISE B The two sides of the triangle are 5 cm and 2 cm long. Can the
third side of the triangle be 7 cm long? Can it be 2 cm long?
Triangle Inequality Theorem
The sum of the lengths of any
two sides of a triangle is greater
than the length of the third side.
a+b >c
b+c >a
a+c >b
Note. If three line segments of different lengths are given, then to determine if a
triangle can be constructed from them, it is enough to check whether the sum of
the lengths of the two shorter ones is greater than the length of the third side.
EXERCISE C Given are line segments with lengths a = 2, b = 5, c = 8, d = 9.
Choose three of them, which can be sides of the same triangle.
A
=
a×h
2
BASIC PROPERTIES OF TRIANGLES
MLR2-1 str. 45
The area of a triangle is equal to half
the product of the length of a side and
the length of the height drawn to the
line containing that side.
45
PROBLEMS
1. a)
Which side of the ABC triangle’s sides is the longest and
which side is the shortest?
b) Which of the KLM triangle’s
angles is the largest and which
is the smallest?
2. a)
The sides of the triangle ABC have the following lengths: |AB| = 17, |BC| = 12,
|AC| = 15. Which angle of this triangle is the largest and which — the smallest?
b) In triangle ABC, angles have measures: |¾ BAC| = 50◦, |¾ CBA| = 85◦. Which side
of this triangle is the longest and which side is the shortest?
3. Check if a triangle can have sides of length:
a) 2 dm, 15 cm, 32 cm
b) 6 mm, 4 cm, 3 cm
√
√
c) 6, 3 2, 7 − 3
√ √
√
d) 8, 18, 50
4. Calculate the areas of the triangles drawn. Assume that the side of the grille is 1.
5. In the figure, AB
and EC are parallel. The
triangle’s ADE area is 35. Calculate the triangles’ ADC, DBC and ABC areas.
6. a)
The sides of√a triangle with an area of
14 are 5, 7 and 4 2. Calculate the length of
the shortest height of this triangle.
√
√
b) The sides of a right-angled triangle are 3, 3 2 and 3 3 long. What is the area
of this triangle?
heights of a triangle are 12, 11 15 and 12 12
long. The perimeter of this
13
triangle is 42. Calculate the lengths of all its sides.
7. The
46
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 46
PYTHAGORAS’ THEOREM AND
INVERSE OF PYTHAGORAS’ THEOREM
Pythagoras’ Theorem
In a right-angled triangle, the sum
of squares of the lengths of catheti is equal
to the square of the length of hypotenuse.
a2 + b2 = c 2
Proof
Let a and b be the lengths of the catheti (commonly: legs) of a right-angled
triangle, and c is the hypotenuse’s length. We will prove that a2 + b2 = c 2 .
We build a square on the hypotenuse c (see first
figure). Then we add three identical right-angled
triangles with sides a, b and c to the three sides of
the square (see the second drawing).
Since the angle of the square with side c and the
adjacent two acute angles of the triangles form a
straight angle, the figure obtained is a quadrilateral, and more precisely — a square with the side
a + b.
The area of this square is: P = (a +b)2 = a2 +2ab +b2
Because the resulting square consists of four triangles and a square with side c, its area can also be
calculated as follows:
P = 4 × a × b + c 2 = 2ab + c 2
2
Therefore, we can write the equality:
a2 + 2ab + b2 = 2ab + c 2
Hence: a2 + b2 = c 2
From history
Pythagoras of Samos (572 – 497 B.C.)
lived in Greece at a time when
Buddha taught in India and Confucius in China. Pythagoras was not
only a mathematician, but also a
philosopher. The philosophy school
he founded proclaimed, among others, faith in reincarnation. Pythagoreans believed that the soul of man
can incarnate even into a plant. They
also conducted scientific activities.
It is not known whether Pythagoras’ theorem was proved for the first
time by Pythagoras himself or by
any of his students. It is certain,
however, that it was known before,
because examples of its use have already been found in Egyptian papyri.
PYTHAGORAS’ THEOREM AND INVERSE OF PYTHAGORAS’ THEOREM
MLR2-1 str. 47
47
Good to know!
Here’s another wording for Pythagorean Theorem:
In a right-angled triangle, the sum of areas of
the squares built on the legs is equal to the
area of the square built on the hypotenuse.
The theorem in this version can be proved
by dividing the two smaller squares into such
parts that they can be made into the larger square. There are many such proofs of
Pythagoras’ theorem (squares can be divided
in various ways). One of them is shown in
the picture beside. A book The Pythagorean
Proposition by E. S. Loomis is a collection of
370 (sic!) different proofs.
EXAMPLE
The size of the TV screen is determined by giving the length of
its diagonal (in inches). What dimensions in centimeters does the 24-inch screen
have if the ratio of its width to height is 4 : 3? (1 inch = 2,54 cm)
We make an auxiliary drawing; 4 : 3 ratio means that if for a
certain x the screen’s width is 4x, then the screen’s height
is 3x.
(4x)2 + (3x)2 = 242
16x 2 + 9x 2 = 242
25x 2 = 242
242
x 2 = 25
24
x= 5
Screen’s height = 3 ×
Screen’s width = 4 ×
We use the Pythagorean theorem (we must remember that
we get the result in inches).
The letter x means the length of a certain line segment,
i.e. it is a positive number, so a negative solution to the
quadratic equation (− 24
) we do not take into account.
5
24
24
inches = 3 ×
× 2,54 cm ≈ 36,6 cm
5
5
24
× 2,54 cm ≈ 48,8 cm
5
Ans. The TV screen has dimensions (approximately) of 48,8 cm × 36,6 cm.
PROBLEM
One of the legs of a right-angled triangle is twice as long as the other.
The hypotenuse has a length of 10. Calculate the legs’ lengths of this triangle.
48
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 48
EXERCISE A The triangle in the picture is equilateral.
The height divides this triangle into two right-angled triangles.
a) Write the equality resulting from Pythagorean Theorem
for one of these triangles and solve it for h.
b) What is the area of an equilateral triangle with side a?
EXERCISE B Triangle in the drawing is a right-angled
isosceles triangle. Write the equality resulting from Pythagorean Theorem for this triangle and solve it for b.
The formulas for calculating the height and area of an equilateral triangle
and the length of the diagonal of a square can always be easily obtained
using Pythagoras’ Theorem.
√
a 3
h =
2
√
a2 3
A=
4
√
d =a 2
EXERCISE C One diagonal of a diamond divides it into two equilateral triangles. What is the ratio of the diagonals’ lengths of this diamond?
EXERCISE D The sentences given are: implication and reverse implication.
Which of these sentences are true?
a) If one of the angles of a triangle is right, then two angles of this triangle are
acute.
If two angles of a triangle are acute, then one of the angles of this triangle is
right.
b) If the triangle is equilateral, then each angle of this triangle is 60◦.
If each triangle’s angle is 60◦, then the triangle is equilateral.
Note that the Pythagorean theorem can be formulated in the form of the
following implication:
If
the triangle is right-angled,
then
the sum of squares of lengths of the two shorter sides
is equal to the square of length of the third side.
We will now substantiate that the reverse implication is also true.
PYTHAGORAS’ THEOREM AND INVERSE OF PYTHAGORAS’ THEOREM
MLR2-1 str. 49
49
Inverse Pythagoras’ theorem
If the sum of squares of lengths of two sides of a triangle
is equal to the square of length of the third side,
then this triangle is a right-angled triangle.
Proof
Let’s assume that the sides’ length of the triangle are a, b and c and that
a2 + b2 = c 2 . Let α be the angle between sides a and b and a ≥ b. Using the
indirect method, we will prove that α = 90◦.
Suppose α is not a right angle. It can therefore be acute or obtuse.
1. Assume that angle α is an acute angle and a ≥ b. Then we have the situation
as shown in the picture. From Pythagoras’
Theorem we get:
h2 = b2 − x2 and h2 = c 2 − (a − x)2
Hence:
b2 − x2 = c 2 − a2 + 2ax − x2
So:
a2 + b2 = c 2 + 2ax,
where 2ax > 0
Therefrom:
a2 + b2 > c 2
The inequality obtained is contrary to the equality assumed at the beginning:
a2 + b2 = c 2 . So α cannot be an acute angle.
We assumed that a ≥ b. For b > a, the proof would be analogous.
2. Suppose α is an obtuse angle. This situation is illustrated in the figure opposite.
Pythagoras’ Theorem indicates that:
h2 = b2 − x2 and h2 = c 2 − (x + a)2
Hence:
b2 − x2 = c 2 − x2 − 2ax − a2
So:
a2 + b2 = c 2 − 2ax, where 2ax > 0
Therefrom:
a2 + b2 < c 2
This inequality is contrary to the assumption. So α cannot be an obtuse angle.
Since angle α cannot be either acute or obtuse, it must so be the right angle.
50
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 50
PROBLEMS
1. Calculate the length of the side labelled with a letter.
2. Calculate the length of the side labelled with a letter.
3. Which
of the figures in the drawing next
to it has a larger perimeter — the triangle or
the parallelogram? How much bigger?
4. In
an isosceles triangle, the arm is three
times longer than the base. Calculate the ratio of the height going toward the base, to
the length of this base.
5. Write a formula that allows calculating:
a) the length of the side a of the equilateral triangle when the area A of this
triangle is given,
b) the perimeter d of the equilateral triangle when the height h of this triangle is
given,
c) the area A of the equilateral triangle when the height h of this triangle is given,
d) the perimeter d of the equilateral triangle when the area A of the triangle is
given.
Worth remembering!
The isosceles right-angled triangle is half the
square, and the one with 30◦ and 60◦ acute
angles is half the equilateral triangle.
The figures show the relationships between
the side lengths in such triangles.
PYTHAGORAS’ THEOREM AND INVERSE OF PYTHAGORAS’ THEOREM
MLR2-1 str. 51
51
6. Can
a box in the shape of a 30 cm × 1 m × 1,5 m be moved through a squareshaped window with a side of 95 cm? Justify the answer.
7. Check
if the triangle whose sides have the specified lengths is a right-angled
triangle.
√
√ p
√
d) 2 2, 3 + 2, 3 + 6 2
a) 20, 26, 32
√
b) 3 7, 12, 9
√
√
√
c) 3 5, 6 2, 3 13
e) 3 dm, 3,6 dm, 20 cm
f) 1,2 m, 5 dm, 130 cm
8. From
among the line segments drawn, select three of which you can build a
right-angled triangle. (Note. The task has five solutions.)
Good to know!
Three natural numbers that can be the
sides of a right-angled triangle are called
the Pythagorean triple. Here are examples of such triples:
3, 4, 5;
5, 12, 13;
40, 198, 202.
Already 3.500 years ago, the Babylonians
knew many such triples. It turns out that
there are infinitely many of them.
Here is a general method for finding
Pythagorean triples:
We choose positive natural numbers p,
q, such that p > q > 0, and calculate a, b
and c according to the formulas:
a = p2 − q 2
b = 2pq
c = p2 + q 2
The numbers thus obtained meet the
condition:
a2 + b2 = c 2
9. a)
Find some Pythagorean triples using the method described above.
b) Justify that the numbers a, b and c calculated from the formulas given above
meet the condition a2 + b2 = c 2 .
c) Find the Pythagorean triple with the largest number being 20.
d) Find the Pythagorean triple with the largest number being 34.
52
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 52
PROPERTIES OF TRIANGLES
(CONTINUED)
We say about two polygons that they
are congruent if the subsequent sides
of one polygon have the same lengths
as the corresponding sides of the other
polygon and the angles between the corresponding sides have equal measures.
To determine if two triangles are congruent, you do not need to check that
all the relevant sides have equal lengths and that all relevant angles have
equal measures. You can use Rules of Congruence for Triangles, which we
remind below.
Rules of congruence for triangles
SSS (side-side-side) rule
If the sides of one triangle have the same
lengths as the corresponding sides of the other triangle, then the triangles are congruent.
SAS (side-angle-side) rule
If two sides of one triangle have the same
lengths as the corresponding sides of the other triangle and the angles between these sides
have equal measures, then the triangles are
congruent.
ASA (angle-side-angle) rule
If a side of one triangle has the same length
as the side of the other triangle, and the angles of one triangle at this side have measures
equal to the measures of corresponding angles of the other triangle, then the triangles
are congruent.
EXERCISE A Which of the of congruence rules of triangles show that the drawn
triangles are congruent?
By using the rules of congruence of triangles, some geometrical properties
can be proved.
PROPERTIES OF TRIANGLES
MLR2-1 str. 53
(CONTINUED)
53
Let us remind you that the perpendicular bisector of a line segment is
perpendicular to it and passes through its center.
Theorem
A point belongs to the perpendicular bisector
of a line segment if and only if
it is equidistant from its ends.
|P A| = |P B|
Proof
We will show that if point P lies on the perpendicular bisector of line segment
AB, then |P A| = |P B|.
Let S be the center of segment AB and P 6= S. Triangles ASP and BSP have a common side SP . In
addition, |AS| = |SB| and the angle with vertex S in
both triangles is right. Thus, from the SAS rule, these
triangles are congruent. Therefore, |P A| = |P B|.
When P = S, of course also |P A| = |P B|.
Now, we will show that if point K is equidistant from the ends of the line
segment AB, it lies on the perpendicular bisector of this line segment.
Let S be the center of the line segment AB and K 6= S.
Triangles ASK and BSK have corresponding sides
of the same length, so the SSS rule implies that
these triangles are congruent. Therefore, |¾ ASK| =
= |¾ BSK|. Because they are supplementary angles,
so both are right angles. It follows that the line KS is
the perpendicular bisector of line segment AB.
When K = S, then obviously K lies on the perpendicular bisector of line segment AB.
Theorem
In each triangle, the perpendicular bisectors of the sides
intersect at one point.
Proof
Let S be the intersection point of the
perpendicular bisectors of AB and AC
in triangle ABC. From the property of
a perpendicular bisector it follows that
|SA| = |SB| and |SA| = |SC|.
From here |SB| = |SC|, which means
that point S also lies on the perpendicular bisector of BC.
54
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 54
Let us remind you that the bisector of an angle is a ray that divides this
angle into two angles of equal measures.
Theorem
A point belonging to a convex angle
lies on its bisector
if and only if
it is equally distant
from both angles’ arms.
We will use this theorem only for acute, right or obtuse angles, so we will
prove it only for such angles.
Proof
We will show that if point P lies on the bisector of the angle with vertex W
(acute, right or obtuse), then the distances of P from the angle arms are equal.
Let P 6= W , P A ⊥ W A and P B ⊥ W B. Triangles W P A
and W P B are right-angled and have a common W P
side. In addition, the angles adjacent to this side are
the same in both triangles. Thus, from the ASA rule it
follows that these triangles are congruent. Therefore,
|P A| = |P B|.
When P = W , then of course the distances of the P
point from both angle arms are equal (equal to 0).
We will show that if K is such a point of an angle (acute, right or obtuse) with
vertex W that it has equal distances from the angle arms, then this point lies
on the bisector of that angle.
Let K 6= W , KR ⊥ W R, KT ⊥ W T and |KR| = |KT |.
Triangles W T K and W RK each have two corresponding sides of the same length. According to
Pythagoras’ Theorem, |W R| = |W T |. The SSS rule implies that these triangles are congruent. Therefore,
|¾ T W K| = |¾ KW R|, i.e. the K point lies on the bisector of the ¾ T W R angle.
When K = W , then of course K lies on the bisector,
too.
Theorem
In each triangle, the bisectors of the angles intersect at one point.
EXERCISE B Prove the above theorem, imitating the proof of the theorem
about the intersection of the perpendicular bisectors of sides of a triangle.
PROPERTIES OF TRIANGLES
MLR2-1 str. 55
(CONTINUED)
55
Bisector theorem
If in triangle ABC the bisector of the angle
with vertex C intersects side AB
at point D, the ratio |AD| : |BD|
is equal to the ratio |AC| : |BC|.
|AD|
|AC|
=
|BD|
|BC|
Proof
Let in the triangle ABC the bisector of the angle with vertex C intersects side
AB at point D. The distances of point D from the arms of this angle are equal.
They are labelled with a in the figure.
The ADC triangle’s area can be written in two ways.
From the equality |AD| × h = |AC| × a equality
|AD|
|AC|
= a results.
2
2
h
Similarly for the BDC triangle, from equality
|BD| × h
= |BC| × a , |BD| = a follows.
2
2
|BC|
h
Therefore: |AD| = |BD| .
|AC|
|BC|
EXERCISE C Draw any two triangles: acute- and obtuse-angled. In each of them
draw lines containing the heights of the triangle.
If your drawings from Exercise C were made accurately, then on each of
them the drawn lines should intersect at one point. The following statement is true.
Theorem
In each triangle, lines containing heights
intersect at one point.
Proof
Let ABC be any triangle. Drawing through each of
its vertices a line parallel to the
opposite side, we will obtain a triangle
A′ B ′ C ′ (see figure).
The quadrilaterals ABA′ C and ABCB ′ are parallelograms, so |A′ C| = |AB| i |B ′ C| = |AB|. Thus
point C is the midpoint of A′ B ′ line segment.
Similarly, it can be seen that point B is the center of A′ C ′ and point
A is the center of B ′ C ′ .
56
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 56
We will now show that the perpendicular bisectors of the triangle
A′ B ′ C ′ contain the heights of the triangle ABC.
Perpendicular bisector
of side B ′ C ′ passes through
point A and is perpendicular to the
line segment BC, because B ′ C ′ k BC.
This perpendicular bisector contains
the height of the triangle ABC going
from vertex A.
Similarly, it can be shown that the perpendicular bisectors of sides of A′ B ′
and A′ C ′ contain the remaining heights
of the triangle ABC going from the vertices C and B.
We already know that the perpendicular bisectors of the triangle A′ B ′ C ′ intersect at one point (see theorem on page 54). This point (labelled in the
drawing with S) is the point of intersection of lines containing the heights of
the triangle ABC.
The line segment connecting the vertex of a triangle with the center of the
opposite side is called a median of the triangle.
EXERCISE D
point?
Draw any triangle and all its medians. Do they intersect at one
The following theorem describes the fourth special point of a triangle.
Theorem
In each triangle the medians intersect at one point and this point
divides each of them in the 2 : 1 ratio.
Proof
Let ABC be any triangle.
Let S be the intersection point of the medians taken from the vertices A and
B (see first figure) and let W be the intersection point of the medians taken
from vertices A and C (see second figure).
PROPERTIES OF TRIANGLES
MLR2-1 str. 57
(CONTINUED)
57
Look at the next drawings. On each of them — by drawing lines parallel to the
sides of the triangle ABC — two triangles adjacent to this triangle were drawn
and the corresponding medians were marked.
The letters x and y as well as w and z indicate the length
of the line segments, to which point S divided the medians. The
letters a and b as well as c and d mean lengths of line segments into which
the W point divided the medians. It can be seen that in both drawings the
shaded figures are parallelograms, so we get equalities:
y = 2x and z = 2w
as well
b = 2a and d = 2c
We have shown that both the S point and the W point divide the medians
going from vertex A in a 2 : 1 ratio, so S = W and S divides each median in this
ratio.
The points described in the above two statements have their names.
The intersection of lines containing
the heights of a triangle is called the
orthocenter of the triangle.
The intersection of the medians of
a triangle is called the center of
gravity of the triangle.
Note that in an equilateral triangle the orthocenter, the center of gravity, the intersection
of perpendicular bisectors and the intersection of bisectors of angles is the same point
that divides each height in the ratio of 1 : 2.
Note. If the triangle is not equilateral, the orthocenter, center of gravity, intersection of perpendicular bisectors and intersection of bisectors of
angles are four different points.
58
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 58
EXERCISE E
Perform the following steps:
1. Draw any triangle (rather large) on a
cardboard and cut it out. Hang the triangle on the hook (select the place of
attachment at some distance from vertices). Hang the weighted thread on the
same hook. Mark the line along the thread
on the triangle.
2. Choose another point to hang the triangle and do the same.
3. Find the intersection of the lines drawn.
Check (by drawing the medians) that this
point is the center of gravity of the triangle.
Good to know!
The concept of center of gravity should be familiar to you from physics
lessons. Gravity is the force acting on every point of the physical body. The
resultant of these forces is applied at a point called the center of gravity.
So, if we will cut a triangle out of cardboard, determine its center of gravity
and attach a thread at this point, then after hanging the triangle on it, it
should hang horizontally and remain in balance.
PROBLEMS
1. Find a triangle in the drawing that is congruent with the shaded triangle.
2. In the triangle ABC in the drawing next
to, points D, E, F are the centers of the sides.
The sides of the DEF triangle are parallel to
the respective sides of the ABC triangle. Justify that the triangles AED, EBF, DEF and
DFC are congruent.
PROPERTIES OF TRIANGLES
MLR2-1 str. 59
(CONTINUED)
59
3. The ABCD quadrilateral in the picture is a parallelogram. Show that the triangle
ABF is congruent with the triangle CDE and that the triangle AED is congruent
with the triangle CFB.
4. The two squares are located as shown below. Show that angle α is right.
5. In triangle ABC
two sides have lengths |AB| = 12 and |BC| = 16. Perpendicular
bisectors of these sides intersect at point S. The diagonal SB of the quadrilateral
ABCS is 10. Calculate the area of this quadrilateral.
6. Point D in the drawing is the intersection point of the bisectors of the two
angles of the triangle ABC. Calculate the angle’s α measure.
7. In triangle ABC, bisectors of angles with vertices A and C
intersect the sides of
the triangle at points E and D, respectively, and intersect at point F. Show that:
a) |AD| = |DF |
|AC|
|CF |
b) |AF | × |DF | = |AD|
|F E| × |CF |
c) |DF | + |BD| = |BE|
|CE|
8. The heights of the equilateral triangle ABC
|CF |
|AC|
|CE|
intersect at point D.
a) Calculate the length of segment AD if it is known that |AB| = 5.
b) Calculate the distance of point D from the line BC if |BC| = 10.
c) Calculate the distance of point D from the side AB if |DC| = 12.
d) Calculate the area of the BCD triangle if |AD| = 20.
60
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 60
9. In the right-angled triangle with sides 6, 8, 10 long the median and the angle
bisector are taken from the right angle’s vertex. They divided the hypotenuse of
the triangle into three line segments. Calculate the lengths of these line segments.
10. In an isosceles triangle, the base has a length of 6 and the length of the arms
is 9. Calculate the distance of the center of gravity of this triangle from its base.
11. Justify
that the medians of a triangle divide it into six triangles with equal
areas.
12. Line segments AE and CD in the drawing are the medians of triangle ABC.
The F point is the center of gravity of this triangle, and the G and H points are the
centers of the CF and AF line segments. Show that |GE| = |DH|.
13. Point D in the figure is the orthocenter of the triangle ABC. Calculate the
angle’s α measure.
14. a)
Prove that in the parallelogram ABCD:
|AC|2 + |BD|2 = 2|AB|2 + 2|AD|2
b) Prove that in the KLM triangle in which |KL| = √
m, |KM| = l and |LM| = k, the
1
length of the median from vertex K to side LM is 2 2m2 + 2l 2 − k2 . Determine the
lengths of the remaining medians of this triangle.
c) Let S be the center of gravity of the KLM triangle. Justify that the sum of
squares of distances of S from the vertices of the triangle is 3 times smaller than
the sum of squares of lengths of the sides of this triangle.
PROPERTIES OF TRIANGLES
MLR2-1 str. 61
(CONTINUED)
61
PROPERTIES OF QUADRILATERALS
We have already discussed the properties of angles in some quadrilaterals
(see page 42). Now we will remind other properties of quadrilaterals.
EXERCISE A Let’s assume that the side of the grille is 1. Calculate the areas of
the figures drawn.
EXERCISE B Draw a square, a rectangle (which is not a square), a rhombus
(which is not a square) and a parallelogram (which is not a rhombus or rectangle). Draw the diagonals of these quadrilaterals. Consider what can be said
about the diagonals of each of the figures drawn.
The square is a quadrilateral whose all angles are right and all sides have
equal lengths.
The area of the square is equal to
the square of the length of its side.
The diagonals of the square have
equal lengths, they intersect in half
and are perpendicular.
A = a2
A rectangle is a quadrilateral whose all angles are right.
The area of a rectangle is equal to
the product of the lengths of its
two adjacent sides.
The diagonals of a rectangle are
equal in length and intersect in the
midpoint.
A=a×b
62
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 62
A rhombus is a quadrilateral whose all sides have equal lengths.
The area of a rhombus is equal to
half the product of the lengths of
its diagonals.
A=
The rhombus’ diagonals intersect
in the midpoint and are perpendicular. The diagonal divides the
rhombus’ angle into two parts of
equal measure.
e×f
2
Note. The square is a rhombus, so its area can also be calculated using the
rhombus’ area formula.
A parallelogram is a quadrilateral that has two pairs of parallel sides.
The area of a parallelogram is equal
to the product of the length of its
side by the length of the hight perpendicular to this side.
The diagonals of a parallelogram intersect in the midpoint.
A=a×h
Note. The rhombus is a parallelogram, so its area can also be calculated using the
formula for the parallelogram’s area.
A trapezoid is a quadrilateral that has at least one pair of parallel sides.
The trapezoid’s area is equal to half
the product of the sum of lengths
of its bases, and the length of its
height.
A=
PROPERTIES OF QUADRILATERALS
MLR2-1 str. 63
In the isosceles trapezium, which is
not a parallelogram, the diagonals
have equal lengths.
(a + b) × h
2
63
The rhombus’ diagonals have lengths 3 and 5. What is the height
of this rhombus?
EXAMPLE
The rhombus’ properties imply that the ABO triangle is
right-angled, and its legs are 32 and 52 long (each of them
is half of the corresponding diagonal).
2 2
5
3
a2 = 2 + 2
We calculate the side’s length of the rhombus using
Pythagorean Theorem for triangle ABO.
34
a2 = 4
a=
√
34
2
We cannot take into account the negative solution of the
quadratic equation, because this is the length of side.
3×5
= a×h
2
√
15
34
=
×h
2
2
√
15
15 34
h= √
= 34
34
We present the rhombus’ area in two ways and write the
appropriate equality.
Ans. The height of the rhombus is
√
15 34
, i.e. about 2,6.
34
PROBLEM
The rhombus’ area is 32 and one of the diagonals is 4. Calculate the
height of this rhombus.
The properties of diagonals of a square, rectangle, rhombus and parallelogram mentioned above can be formulated using equivalence. Here is an
example of such an equivalence:
Theorem
A quadrilateral is a parallelogram if and only if
its diagonals intersect at the midpoint of each.
Proof
We will show that if the ABCD quadrilateral is a parallelogram, then its diagonals intersect in the midpoint.
Let P be the intersection point of the diagonals of the ABCD parallelogram. The ASA
rule tells that the triangles ABP and CDP are
congruent because |AB| = |CD| and the respective angles of these triangles are equal.
From here: |AP | = |P C| and |BP | = |P D|.
Thus, point P is the center of the diagonal AC and center of the diagonal BD.
64
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 64
We will now show that if KLMN is a quadrilateral whose diagonals intersect in
the midpoint, it is a parallelogram.
Let S be the intersection point of the KLMN’s
diagonal. The triangles KLS and MNS are
congruent (rule SAS).
Therefor: |¾ KLS| = |¾ SNM|. So, KL k MN.
Similarly, it can be shown that KN k ML.
It follows that the KLMN quadrilateral has two pairs of parallel sides, it is so
a parallelogram.
EXERCISE C Draw two line segments of different lengths, but intersecting in
the midpoint of each. Draw a quadrilateral whose diagonals are these line
segments. Which one of the implications proved above determines what quadrilateral it is?
It is known that every parallelogram
is a trapezoid, every rectangle — a
parallelogram, etc.
EXERCISE D Using the trapezoid’s formula, calculate the area of:
a) a square with side of length a,
b) a rectangle with sides of lengths a
and b,
c) a parallelogram with side a and
height h perpendicular to this side.
PROBLEMS
1. Calculate the areas of the figures shown in the drawings.
PROPERTIES OF QUADRILATERALS
MLR2-1 str. 65
65
2. The
drawn lines are parallel. Which figure has the largest area? Which figures
have equal areas?
3. Write the formula to calculate:
a) the square’s area with a given diagonal d,
b) the perimeter of a square with a given area A,
c) the rhombus’ area with given perimeter P and height h,
d) the rhombus’ perimeter with diagonals e and f given,
e) the area of an isosceles trapezoid with given bases a, b (a > b) and arm r ,
f) the perimeter of an isosceles trapezoid with bases a and b (a > b) and area A.
4. In
a right-angled trapezoid with a perimeter of 8, the shorter base and height
are equal in length, the base length’s difference is 2. Calculate the trapezoid’s base
lengths.
Calculate the area of a rhombus in which one of the angles is 45◦and the side
is of length a.
5. a)
b) Calculate the isosceles trapezoid’s area in which one of the angles is 60◦, shorter
base is of length a and the arm is 2a long.
c) Calculate the perimeter of a parallelogram with heights a and b if one of its
angles has measure 30◦.
6. Read the Curiosity.
Curiosity
A-format paper sheets (e.g. A4 —
used for printers, A5 — notebook
pages) have the following property: if
we draw from the top a line segment
at an angle of 45◦ (see figure), this
line segment has the same length as
the longer side of the sheet.
a) Calculate the ratio of the length of A-format paper to its width.
b) A-format page is divided into an isosceles right-angled triangle and a trapezoid.
How many times the trapezoid’s area is larger than the triangle’s area?
66
FIGURES ON THE PLANE. PART 1
MLR2-1 str. 66
7. The
figures show the carousel at rest and in motion. How high up is the chair
when the carousel is spinning?
8. Look
at the drawing. A rope was attached to the branch and a 60 cm long
board was attached to it. The swing hangs 70 cm above the ground. When a shorter
board was attached instead of this board, the swing hung 60 cm above the ground.
Calculate the length of the shorter board.
9. Segment
MN connects the centers of the opposite sides of the parallelogram
ABCD. Show that the segment MN and the diagonal BD intersect in half their
lengths.
PROPERTIES OF QUADRILATERALS
MLR2-1 str. 67
67
MLR2-1 str. 68
Functions
Sound is a physical phenomenon caused by vibrations. Thus, the magnitude
of this phenomenon can be measured by the energy of these vibrations.
However, it turns out that sound of twice the energy is not perceived
by people as twice as loud. When we want to compare sound levels according
to how the human ear perceives them, we use the logarithmic function.
Polynomial functions
Polynomial inequalities
Polynomial functions (continued)
(continued)
Polynomial inequalities
Exponential and logarithmic functions
Exponential and logarithmic equations
and logarithmic functions
MLR2-1 str. 69
Applications of exponential
POLYNOMIAL FUNCTIONS
Below are formulas and fragments of graphs of several functions. The formula of each of them has the form y = W (x), where W (x) is a polynomial.
These types of functions are called polynomial functions. The domain of a
polynomial function is the set of real numbers.
y = −x3 + 2x2 + 3x − 3
y = −x4 + x3 + 2x2 + 1
y = x (x2 + 8x + 15)(x2 − 6x + 8)
50
EXERCISE A If you have a graphing calculator or a computer with a suitable
program, use them to make graphs of any polynomial functions.
Note that linear and quadratic functions (which were discussed in the first
class) are also polynomial functions.
Below we recall their properties.
The function given by the y = ax + b formula is a linear function. Its graph
is a line. When a > 0, the function is increasing, when a < 0, it is decreasing,
and when a = 0, it is constant.
Graphs of functions
of the type y = ax + b
with the same coefficient a
are parallel lines.
70
Graphs of functions
of the type y = ax + b
with the same coefficient b
intersect at (0, b).
FUNCTIONS
MLR2-1 str. 70
The function defined by formula y = ax2 +bx+c, where a 6= 0, is a quadratic
function. Its graph is a parabola with arms pointing up when a > 0, or
down when a < 0. The number of zeros of such a function depends on the
value of ”delta”, i.e. the expression ∆ = b2 − 4ac.
If ∆ > 0, the function has two zeros:
If ∆ = 0, the function has one zero:
√
x1 = −b − ∆ ,
2a
√
x2 = −b + ∆ .
2a
x0 = −b .
2a
If ∆ < 0, the function has no zeros.
The formula of a quadratic function can
be written in the standard form, and if
the function has zeros, it can also be in
the factored form.
Standard form: y = ax2 + bx + c
Vertex form: y = a(x − p)2 + q
Factored form: y = a(x − x1 )(x − x2 )
When the quadratic function has one zero x0 ,
the function formula can be written in the
form y = a(x − x0 )2 .
p=− b
2a
q=
−∆
4a
p = x1 + x2
2
q = f (p)
We will now discuss some properties of the functions in the form y = axn ,
where n ∈ Ž+ .
EXERCISE B
Look at the function graphs below.
a) Determine the zero of each of these functions.
b) Determine the monotonicity of each of these functions.
c) Which of the graphs is symmetrical respective the y-axis, and which —
relative to the origin of the coordinate system?
POLYNOMIAL FUNCTIONS
MLR2-1 str. 71
71
Graphs of functions of the type y = axn (n ∈ Ž+ ) pass through the origin
of the coordinate system. Also:
If n is an even number, then the
graph of the function y = axn has
an axis of symmetry — it is the
y axis. Depending on the value of
the coefficient a, the function can
only take non-negative values (for
a > 0) or only non-positive values
(for a < 0).
If n is an odd number, then the
graph of y = axn has the center of
symmetry — it is the system’s origin. Depending on the value of the
coefficient a, the function can be
increasing (for a > 0) or decreasing
(for a < 0).
EXERCISE C Sketch the graphs of the given functions and determine if these
functions have zeros. If the function has zeros, specify their number and signs.
f (x) = 5x24 − 6
g(x) = −7x109 + 3
h(x) = −3x50 − 4
In the frame below, we remind you of the basic information on transforming function graphs.
When the graph of the function y = f (x) is shifted by vector [p, q], we
get the graph of the function y = f (x − p) + q.
When the graph of the function y = f (x) is reflected symmetrically
about the x axis, we get the graph of the function y = −f (x).
When the graph of the function y = f (x) is reflected symmetrically
about the y-axis, we get the graph of the function y = f (−x).
72
FUNCTIONS
MLR2-1 str. 72
Note that by shifting or symmetrically reflecting function graphs of the
type y = axn , we get graphs of other polynomial functions. Examples are:
After moving the function graph:
After moving the function graph:
f1 (x) = 2 x3
f2 (x) = −3x4
by vector [−1, 4] we get the graph of
the function:
by vector [2, −5] we get the graph of the
function:
3
g1 (x) = 2 (x + 1) + 4
g2 (x) = −3(x − 2) − 5
3
3
After moving the function graph:
f3 (x) = 3x
3
4
After moving the function graph:
f4 (x) = − 3 x6
4
by vector [0, 4] we get the graph of
the function:
g3 (x) = 3x3 + 4
by vector [−2, 0] we get the graph of the
function:
6
g4 (x) = − 3 (x + 2)
When we reflect this graph symmetrically relative to the x-axis, we get the
graph of the function h3 (x) = −3x3 −4.
When we reflect this graph symmetrically relative to the y-axis, we get the
6
graph of the function h4 (x) = − 43 (−x + 2) .
4
We discuss other properties of polynomial functions in one of the next chapters. .
POLYNOMIAL FUNCTIONS
MLR2-1 str. 73
73
PROBLEMS
1. Match formulas to graphs.
a)
b)
y =x+1
y =x+3
y = −x + 1
y = −x − 2
y = 1x + 1
y = −2x + 3
y = −1x + 1
2
y = 3x + 2
2
2. Find the formula for the linear function that satisfies the condition:
a) The function graph passes through points (5, −2) and (−4, 1).
b) The zero of the function is −2 and the graph of this function intersects the y
axis at (0, 6).
3. Match formulas to graphs.
a)
c)
y = 1 (x − 2)(x + 5)
2
y = 1 (x + 2)(x − 5)
2
y = x2 − 2x − 4
y = − 1 (x + 2)(x − 5)
y = x2 + 2x + 4
2
y = 1 x2 − 2x + 4
2
b)
d)
y = −2(x − 2)2 + 5
y = −2(x + 2)2 + 5
y = − 1 (x − 2)2 + 5
2
y = −(x + 2)(x − 3)
y = −(x + 4)2 + 8
y = −x2 + 6x − 5
74
FUNCTIONS
MLR2-1 str. 74
4. a) Find the formula of the linear function f whose zero is 3 smaller than the
zero of the function g(x) = − 4 x + 3, but graphs of functions f and g intersect the
y-axis at the same point.
5
b) Linear function f takes positive values only for x ∈ −∞ ; −4 and its values for
arguments −10 and 4 differ by 3. Find the formula for this function.
5. Graphs of the following functions are drawn below:
f (x) = −3x4
h(x) = 1 x3
g(x) = −2x5
k(x) =
4
√ 4
2x
Match the graphs to the function formulas.
6. How
many intersection points can the graphs of the functions f (x) = xn and
g(x) = xm have, where m, n ∈ Ž+ and n 6= m? Consider different cases.
7. Show that the graph of function
metry.
f (x) =


 (x − 3)3

 −x3
for x ≥ 3
2
for x < 3
has an axis of sym-
2
POLYNOMIAL INEQUALITIES
EXERCISE
Using the graph, give three numbers satisfying the given inequality.
a) (x + 2)(−x2 + 10x − 21) > 0
POLYNOMIAL INEQUALITIES
MLR2-1 str. 75
b) (x + 2)(−x2 + 10x − 21) ≤ 0
75
You can now solve first degree inequalities and quadratic inequalities.
These skills can be used to solve some of the higher-degree inequalities.
A simple property will be useful — the product of two numbers is a negative number when factors have different signs, and a positive number
when they have the same signs.
a×b <0
⇐
⇒
a>0
or
b<0
a<0
b>0
a×b >0
⇐
⇒
a>0
or
b>0
a<0
b<0
We will now show you how to solve the inequality:
(x − 3)(x2 + x − 2) < 0
This inequality is satisfied when the values of x − 3 and x2 + x − 2 have
opposite signs, i.e. if and only if:
(
(
x−3 >0
x−3< 0
or
2
x +x−2<0
x2 + x − 2 > 0
To find solutions to these inequality systems, it is most convenient to
sketch in one coordinate system graphs of the functions y = x − 3 and
y = x2 + x − 2 , as in the figure below.
Note. The graphs do not need to be drawn very accurately; it is only important
to be able to read from them for which arguments the function takes negative
values and for which positive (so it is enough to determine zeros, determine
direction of the line and heading of the parabola’s arms).
In the figure, plus and minus signs indicate in which intervals these functions assume positive and in which negative values.
From the drawing we can read that the values of the functions y = x − 3
and y = x2 + x − 2 have opposite signs in the interval −∞ ; −2 , and also in
the interval of 1 ; 3 . Thus, the inequality under consideration is satisfied
for x ∈ −∞ ; −2 ∪ 1 ; 3 .
76
FUNCTIONS
MLR2-1 str. 76
EXAMPLE
Solve the inequality.
4x 3 + 2x 2 − 1 ≥ x 3 + 27x + 17
3x 3 + 2x 2 − 27x − 18 ≥ 0
(3x + 2)(x 2 − 9) ≥ 0
x2 − 9 = 0
3x + 2 = 0
2
x = −3
We transform inequality to the form
W (x) ≥ 0, where W (x) is a polynomial.
We decompose the polynomial into two
or more second-degree factors.
We find the zeros of the functions
f (x) = 3x + 2 and g(x) = x 2 − 9.
x = 3 or x = −3
We sketch graphs of the function f (x) = 3x +2
and g(x) = x 2 − 9 and we mark arguments the
functions have positive/negative values for.
D
E D
2
x ∈ −3 ; − 3 ∪ 3 ; +∞
PROBLEM
From the graph, we read intervals in which
the function values have the same sign or
are equal to 0.
Solve the inequality: x3 + x2 + x + 2 < 5x + 6.
PROBLEMS
1. Each drawing presents fragments of graphs of two polynomial functions of the
first and second degree. Give sets of solutions to the inequalities written under the
drawings.
POLYNOMIAL INEQUALITIES
MLR2-1 str. 77
77
2. Using the graphs of functions f
and g, give the solution of the inequality written
under the figure.
x
a) f (x) = − 2 + 1
g(x) = −x2 + 7x − 10
x
− 2 + 1 (−x2 + 7x − 10) ≥ 0
b) f (x) = 2x + 4
1
c) f (x) = − 5 (x2 + x − 12)
1
g(x) = 2 x2 + x − 4
(2x + 4)
1 2
x
2
+x−4 ≤0
g(x) = 2x − x2
1
− 5 (x2 + x − 12)(2x − x2 ) ≤ 0
3. The
figure presents graphs of functions f and g. Using the formulas, write the
fourth-degree polynomial inequality so that the set given below the figure be the
set of its solutions.
a) f (x) =
1 2
x
4
1
+ 2x − 2
g(x) = −x2 − 4x
b) f (x) = x2 − 8x + 15
g(x) = −x2 + 4
c) f (x) = x2 − 4x + 3
g(x) = − 31 x2 − 13 x + 2
4. Solve inequality (first decompose the polynomial into factors).
a)
1 3
x
2
− 5x2 < 0
e) −4x3 + 3x2 + 4x − 3 > 0
b) 2x4 − x3 ≥ 0
f) −2x3 + x2 + 18x − 9 ≤ 0
c) x3 − x2 − 6x < 0
g) x3 + 5x2 + 8x + 40 ≤ 0
d) 2x7 − 3x6 − 2x5 > 0
h) x4 + x3 − 8x − 8 ≥ 0
5. a)
For what value of p is p3 + p2 − 9p greater than 9?
b) For what natural numbers n is the number 2n3 + 3n2 + 4n − 5 greater than the
number 3n3 − 2n2 + 15?
6. For
2
which
the set of solutions to inequality (x + b)(x + x − 1) > 0 is the
√ value b
5
−
1
interval
; +∞ ?
2
78
FUNCTIONS
MLR2-1 str. 78
POLYNOMIAL FUNCTIONS
(CONTINUED)
Let us remind you that a polynomial of degree n has at most n roots, and
if n is odd, it has at least one root. It follows that:
• The polynomial function y = W (x) , where W (x) is a polynomial of degree n,
has no more than n zeros.
• If W (x) is an odd-degree polynomial, then the function of the form y = W (x)
has at least one zero.
If W (x) is an even-degree polynomial, then the function of the form y = W (x) may
not have zeros.
EXERCISE A
Formula of the given function can be written as:
y = an xn + . . . + a1 x + a0
Give the sign of the coefficient an . Determine the largest zero of the given
function and check for several arguments larger than this zero whether the
function assumes positive or negative values for these arguments.
a) f (x) = 5(x − 2)(x + 3)(x − 5)
b) f (x) = −2(x + 1)(x − 1)(x + 4)
Below are fragments of graphs of several polynomial functions of the form
y = an xn + an−1 xn−1 + . . . + a1 x + a0 and all zeros are marked. On the left are
functions for which an is a positive number, and on the right are functions
for which an is a negative number.
an > 0
If an > 0 and polynomial function
n
n−1
y = an x + an−1 x
+ ... + a1 x + a0
an < 0
If an < 0 and polynomial function
y = an xn + an−1 xn−1 + ... + a1 x + a0
has zeros, then for arguments larger
than all zeros the function values are
positive.
has zeros, then for arguments larger
than all zeros the function values are
negative.
If this function does not have zeros,
then all its values are positive.
If this function does not have zeros,
then all its values are negative.
For arguments smaller than all zeros, the function values can be positive or
negative both when an > 0 and when an < 0.
POLYNOMIAL FUNCTIONS
(CONTINUED)
MLR2-1 str. 79
79
EXERCISE B Each drawing presents a fragment of a graph of function of the
type y = (x + 2)m (x − 3)n . Determine how the sign of the value of the function
changes when its graph is passing through points (−2, 0) and (3, 0), depending
on whether m and n are even or odd numbers.
EXERCISE C What values does the given function take — positive or negative
— for arguments greater than 7, and what ones for those less than 7?
a) y = (x − 7)16
b) y = (x − 7)19
If number a is a root of polynomial W (x), then the graph of function
y = W (x) passes through point (a, 0). After having passed through this
point, the graph can remain on the same side of the x-axis or go to the
other side. It depends on the multiplicity of the root a:
• If a is an even-multiplicity root,
the graph after having passed
through point (a, 0) remains on
the same side of the x-axis (the
sign of the function’s value does
not change).
• If a is an odd-multiplicity root,
then the graph after having
passed through point (a, 0) goes
to the other side of the x-axis
(the sign of the function’s value
changes).
80
FUNCTIONS
MLR2-1 str. 80
EXERCISE D Suppose numbers −1 and 2 are the only roots of the polynomial
W (x). Sketch what the graph of y = W (x) might look like if:
a) both roots of this polynomial are of even multiplicity,
b) number −1 is a simple root and number 2 a trifold root.
PROBLEMS
1. The
figure beside presents four
graphs of third-degree polynomial
functions. For which of these functions is the coefficient at the highest power positive?
2. Graphs
of the functions f and g are drawn below. How should you move the
graph of each of these functions to obtain a graph that:
a) does not intersect the x-axis,
b) has one common point with the x-axis,
c) has five common points with the x-axis,
d) has six common points with the x-axis?
3. The curve in the
drawing is a fragment of the graph of a polynomial function.
Determine the lowest possible degree of the corresponding polynomial.
POLYNOMIAL FUNCTIONS
(CONTINUED)
MLR2-1 str. 81
81
4. Next
to it is a fragment of graph of a polynomial function, in which all its zeros are visible.
Read the roots of the polynomial W (x) from the
graph and determine which of them are of even
multiplicity. How many evenfold roots has the
polynomial y = [W (x)]2 ?
5. Give an example of a polynomial function with
zeros 1, 3 and 5, which assumes:
a) only non-negative values,
b) negative values only for x ∈ 3 ; 5 ∪ 5 ; +∞ ,
c) negative values only for x ∈ −∞ ; 1 .
Curiosity
It is easy to see that the graph of function of the type y = ax3 has a center of
symmetry (it is the origin of the coordinate system). It would seem that graphs
of other third-degree polynomial functions are not as regular. It turns out
however, that each graph of a polynomial function y = ax3 + bx2 + cx + d,
where a 6= 0, has the center of symmetry.
The center of symmetry of this function
graph is this point on the graph whose
b
first coordinate is − 3a
.
6. Read the Curiosity. Only two of the figures below present fragments of plots of
polynomial functions of the third degree. Which ones?
7. Read the Curiosity.
a) Calculate the coordinates of the center of symmetry of the graph of the function
y = −x3 + 3x2 + 7x + 11.
b) What condition must a, b, c and d coefficients fulfill in the formula of function
y = ax3 + bx2 + cx + d so that the center of symmetry of its graph lie on the y-axis?
c) What condition must meet the coefficients a, b and c so that the center of
symmetry of the graph of the function y = ax3 + bx2 + cx lie on the x-axis?
82
FUNCTIONS
MLR2-1 str. 82
POLYNOMIAL INEQUALITIES
(CONTINUED)
EXERCISE A Next to it is a graph
of a function y = W (x). Read
from this graph the solution of
the inequality W (x) ≤ 0.
Using the graph of the polynomial function y = W (x), you can easily provide solutions for the inequality W (x) ≥ 0, W (x) < 0, etc.
If you do not have a computer or graphing calculator available, drawing a
graph of a polynomial function can be very difficult. But using the properties of polynomial functions discussed in the previous chapter, we can
sketch a drawing illustrating how the sign of the value of the function
y = W (x), where W (x) = an xn + . . . + a1 x + a0 .
Here’s how to sketch such a drawing.
• First, we find the roots of the polynomial W (x) and determine their multiplicity, and then mark these roots on the x-axis (even-multiplicity ones can be
marked with a dot, and odd-multiplicity ones with a dash).
• We determine what sign has the polynomial’s W (x) coefficient an at the highest variable power. We start drawing the graph from the right; when an > 0
— we start above the x-axis, when an < 0 — we start below the x-axis.
• We draw a curve passing through the points marked on the axis. If the point
on the axis corresponds to the even-multiplicity root — the curve remains
on the same side of the axis. If the point corresponds to the odd-multiplicity
root — the curve goes to the other side of the x-axis.
Note. It should be remembered that the curve we draw as described above is not
the graph of the given polynomial function. This curve only illustrates how the
sign of the value of the function changes.
POLYNOMIAL INEQUALITIES
MLR2-1 str. 83
(CONTINUED)
83
EXERCISE B The numbers a = −3, b = 0 and c = 2 are the only roots of the
polynomial W (x). Knowing that a is a double root, b — trifold root, and c — a
simple root, sketch a graph illustrating how the sign of the function y = W (x)
changes when the coefficient at the highest power of the variable x is a negative
number.
EXAMPLE
Solve the inequality.
x 6 + 2x 5 − 4x 4 − 8x 3 ≤ 0
x 6 + 2x 5 − 4x 4 − 8x 3 =
= x 3 (x 3 + 2x 2 − 4x − 8) =
We decompose the polynomial into factors to
find its roots and determine their multiplicity.
= x 3 [x 2 (x + 2) − 4(x + 2)] =
= x 3 (x + 2)(x 2 − 4) = x 3 (x + 2)2 (x − 2)
x 3 (x + 2)2 (x − 2) = 0
x =0
x = −2
x =2
trifold
root
double
root
simple
root
We find the roots of the polynomial and determine their multiplicity.
We sketch a chart of change of the sign of
value of the function y = x 6 + 2x 5 − 4x 4 − 8x 3 ;
sketching begins on the right above the xaxis, because the coefficient at x 6 is positive.
The graph intersects the x-axis at (0, 0) and
(2, 0), and on passing through the point
(−2, 0) it remains on the same side of the xaxis.
From the chart we read the solution for the
inequality.
D
E n o
x ∈ 0 ; 2 ∪ −2
PROBLEM
Solve the inequality: (x3 − x2 )(x3 − x2 − x + 1) ≥ 0.
PROBLEMS
1. Solve the inequality.
3
a) −2(x − 3)2 (x − 1)5 x − 1 ≥ 0
2
2
3
e) x(x − 2) (x + 2) (1 − x) > 0
b) 1 x3 (x + 4)4 (x − 3)2 ≤ 0
2
√
√
√
c) − 5x2 (x − 2)3 (x + 3)5 > 0
g) (2x − 6)5 (6x − 3)5 (2 − x)2 ≥ 0
d) −0,4(x − 7)6 (3x + 1)8 (x − 1)7 < 0
h) −5 (3 − 2x) (x − 3) (1 − x) < 0
84
f) (4 − 5x)3 (x − 11)6 > 0
3
4
2
FUNCTIONS
MLR2-1 str. 84
2. Solve the inequality.
2
a) (x − 13) (x − 25)(x + 3) ≥ 0
c) (3x − 2)4 (2x2 − 4)3 (x + 3)2 < 0
b) −7(3 − x)(x2 + 8)(x2 − 100) ≤ 0
d) (x − 5)2 (2x2 − 8)3 > 0
3. Solve the inequality.
a)
3
5
x2 − 25 (5 − x) ≥ 0
5
3
d) (x2 − 2x − 15)4 (x − 5)3 ≤ 0
7
b) 3(2x − 5) (3 − x) (5 − 2x) < 0
e) (x2 − 9)(x + 3)(1 − x2 ) < 0
c) x3 (8 − 2x2 )(2 − x)(3x − 6) > 0
f) (x2 − 1)(2x2 + x − 1)(3x2 + x − 2) ≥ 0
4. Solve the inequality.
a) (x2 − 3x − 10)(x − 5)3 (x2 + 4x + 4) ≥ 0
c) 1 x(x2 + 2x − 3)(x2 + 6x + 9)(x2 − x) ≤ 0
b) − 2 (x2 − 4)(x − 2)(x2 − 1)(x2 − x − 2) > 0
3
d) −3x2 (x2 − 9)(x2 − 6x + 9)(x2 − 3x) < 0
2
EXPONENTIAL AND LOGARITHMIC
FUNCTIONS
You already know that for every real number x the number 2x is specified.
So, you can consider the function y = 2x , whose domain is the set ’.
EXERCISE A Calculate the values of the function y = 2x for the arguments:
−3, −2, −1, 0, 1 , 1, 2, 3 and mark in the coordinate system the corresponding
2
points of the function graph. Sketch the graph of this function.
The figure shows the graph of function y = 2x . Note that:
• The function values are positive because
2x > 0 for each real number x.
• The graph does not intersect the x axis, but for smaller arguments the corresponding points on the graph are closer
and closer to this axis. We say that the
x axis is the horizontal asymptote of the
graph of y = 2x .
• The function graph intersects the y axis
at the point with coordinates (0, 1), because 20 = 1.
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
MLR2-1 str. 85
85
EXERCISE B
The diagram beside
x shows
1
. Check
graph of the function y =
2
which of the features described above
has this function.
Each function whose formula can be written in the form y = ax ,
where a > 0, is called an exponential function. The domain of the
exponential function is the set of real numbers.
Note that for a = 1 the exponential function has the form y = 1, so it
is a constant function. Below are graphs of several exponential functions
y = ax , where a 6= 1, and their properties are given.
• The set of function values is the interval 0 ; +∞ .
• The x-axis is the horizontal asymptote of the function graph.
• The function graph intersects the y-axis at the point with (0, 1) coordinates.
• For a > 1 function y = ax is increasing, and for 0 < a < 1 it is decreasing.
Next to, graphs are presented
func of
x
tions f (x) = 2x and g(x) = 21 . Note
that the formula of function g can be
written in the form g(x) = 2−x . So, there
is equality:
f (−x) = g(x)
Therefor
x the graph of function g(x) =
1
= 2
is symmetrical to the graph of
function f (x) = 2x relative to the y-axis.
Because each positive number x can be uniquely assigned the number
log2 x (that is, such an exponent to which 2 should be raised to get x), so
we can consider the function log2 x, whose domain is set ’+ .
86
FUNCTIONS
MLR2-1 str. 86
The figure shows the graph of function y = log3 x. Note that:
• The domain
of the function is the set
0 ; +∞ .
• The function graph does not intersect
the y-axis, but the closer the arguments
are to zero, the more the corresponding
graph points approach the y-axis. We
say that the y-axis is the vertical asymptote of the function graph.
• The function graph intersects the x-axis
at point (1, 0).
EXERCISE C
The next picture shows the graph
of the function y = log 1 x. Which of the proper3
ties given above has this function?
Each function whose formula can be written as y = loga x, where
a > 0 and a 6= 1, is called a logarithmic function. The domain of the
logarithmic function is the interval (0 ; +∞).
Here are examples of graphs of several logarithmic functions and their
properties.
• The set of logarithmic function’s values is the set of real numbers.
• The y axis is the asymptote (vertical) of the function graph.
• The function graph intersects the x axis only at the point (1, 0), i.e. the
only zero of the function is x = 1.
• For a > 1 the function y = loga x is increasing, and for 0 < a < 1 it is
decreasing.
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
MLR2-1 str. 87
87
Using graphs of exponential and logarithmic functions, graphs of other
functions can be obtained.
The shift of the function graph
x
f1 (x) = 3
2
The shift of the function graph
f2 (x) = log3 x
by vector [3, −2] gives the graph of
by vector [−2, −3] gives the graph of
function
function
g2 (x) = log3 (x + 2) − 3
x − 3
g1 (x) = 3
−2
2
The graph of g1 has a horizontal
The graph of g2 has a vertical asymp-
asymptote with equation y = −2.
tote with equation x = −2.
The symmetrical reflection
Performing the symmetrical reflection
of the function graph
x
f3 (x) = 1
of the function graph
3
f4 (x) = log 2 x
3
relative to the x-axis,
relative to the y-axis, and then reflecting
and then to the y-axis
the part of the received graph, which lies
provides the graph of function
−x
g3 (x) = − 1
below the x-axis, relative to this axis,
3
we receive the graph of the function
g4 (x) = | log 2 (−x)|
3
The asymptote of g3 is x-axis.
88
The asymptote of g4 is y-axis.
FUNCTIONS
MLR2-1 str. 88
The concept of logarithm is associated with exponentiation. Similarly,
there is a relationship between the logarithmic function and an exponential function.
Look at the picture beside. It is easy to see that
the x-axis and y-axis are symmetrical to each other
with respect to the line y = x and each point with
coordinates (a, b) is symmetrical relative this line to
the point (b, a). This fact can be used when drawing
graphs of logarithmic functions.
EXERCISE D Mark four points belonging to the graph of function y = 2x . Then
record the coordinates of the points symmetrical to them relative to the line
y = x. Check that they belong to the graph of the function y = log2 x.
Let’s consider the following functions:
f (x) = 2x
g(x) = log2 x
Let the point (p, r ) be an arbitrary point of the f function
graph. So:
r = 2p
Therefrom:
log2 r = p
Therefore, point (r , p) belongs to
the graph of function g. It means
that the graph of function g is
symmetrical about the line y = x
to the graph of function f .
Similarly, the graph of function
y = log 1 x is symmetrical to the
x
2
function graph y = 1
relative
2
to the y = x line.
EXERCISE E
Draw the graph of function y = 2x + 1 and the graph symmetrical
to it with respect to the line y = x. You can get the same graph by moving the
graph of function y = log2 x accordingly. Record the formula of the obtained
function.
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
MLR2-1 str. 89
89
PROBLEMS
1. Each of the drawings below presents graphs of two of the following functions:
x
g(x) = 6
f (x) = 0,6x
h(x) = 2,7x
7
k(x) = 10x
Match the formulas to the graphs.
2. From among the given formulas, find the ones that represent the same function.
a(x) = 10x + 2
c(x) = 100 + 10x
x
g(x) = 10
i(x) = 10x
b(x) = 10x − 2
d(x) = 100 × 10x
h(x) = 100x
j(x) = 102x
3. a)
100
2
For what value a does the graph of function y = ax pass through point (2, 4)?
b) For what value a point (−3, 27) belongs to the graph of function y = ax ?
4. Specify the domain of function:
a) y = log3 (x − 2)
d) y = log(x2 − 3x − 10)
g) y = logx (x + 1)
b) y = log(3 + 2x)
e) y = logx 2
h) y = logx (2 − x)
f) y = logx − 3 5
i) y = log2 |x − 1|
2
c) y = log 1 (x − 9)
2
5. Which of the following formulas represent the same function?
f (x) = 2 log3 x
i(x) = log 1
x
g(x) = log3 2x
j(x) = log3 2 + log3 x
l(x) = − log x
h(x) = 2 + log3 x
k(x) = log3 (2 + x)
m(x) = log3 9x
6. a)
Justify that the formulas y = log5 x2 and y = 2 log5 x do not represent the
same function.
b) Draw the graph of function y = 1 log2 x2 .
2
90
FUNCTIONS
MLR2-1 str. 90
EXPONENTIAL AND LOGARITHMIC
EQUATIONS
A function is said to be injective when it takes different values for any
two different arguments. In other words: this function assigns each value
for only one argument. This means that the following condition is met:
f (x1 ) = f (x2 )
⇐
⇒ x1 = x2
Note that each increasing function and each decreasing function is injective, so this condition is met by exponential and logarithmic functions.
Equation:
Equation:
3x = 9
you can write like this:
log5 x = 2
you can write like this:
3x = 32
log5 x = log5 25
The exponential function is injective,
so:
x=2
The logarithmic function is injective,
so:
x = 25
and the number 2 is the only solution
to the equation under consideration.
and the number 25 is the only solution
to the equation under consideration.
Note that the function y = x2 is not injective. The equation x2 = 9 can be written
in the form x2 = 32 and x = 3 is a root of this equation, but it is not the only
root of this equation.
The equation 3x = 9 is an example of an exponential equation, i.e. one in
which the unknown occurs only in the exponent.
In the equation log5 x = 2 the unknown is a term in logarithm. This is an
example of a logarithmic equation.
EXPONENTIAL AND LOGARITHMIC EQUATIONS
MLR2-1 str. 91
91
Solve the equation.
EXAMPLE 1
1
We write both sides of the equation in the
form of powers with equal bases.
a) 5x = 25
5x = 5−2
We use the condition: a p = a r
x = −2
b) 3x =
√
5
3x = 3
x=
⇐
⇒ p = r.
3
1
5
1
5
PROBLEM
√
3
a) 2x = 2
Solve the equation.
b) 10x = 0,001
When solving exponential equations, properties of powers are sometimes useful, which
we recall next.
√
3
4
a) 2x =
16
√
3 2
2
2x = 24
(am )n = am × n
am × an = am + n
Solve the equation.
EXAMPLE 2
If a > 0 and m, n ∈ ’,
then:
am
an
= am − n
Using properties of powers, we write both
sides of the equation in the form of powers
with identical bases.
2
23
2x = 24
2
2x = 2 3 −4
10
3
2x = 2−
We use the condition: a p = a r
10
x =− 3
⇐
⇒ p = r.
x
27
b) 9x = √
3
2 x
(3 ) =
(33 )
x
1
32
Using properties of powers, we write both
sides of the equation in the form of powers
with identical bases.
1
32x = 33x − 2
2x = 3x −
1
2
We use the condition: a p = a r
1
x= 2
PROBLEM
√
a) 4x = 2
⇐
⇒ p = r.
Solve the equation.
92
b) 8x = 2 × 4x
FUNCTIONS
MLR2-1 str. 92
Solve the equation.
EXAMPLE 3
4x − 6 × 2x − 16 = 0
All powers we write in the form of powers
with the same bases.
(22 )x − 6 × 2x − 16 = 0
x
(22 ) = (2x )2
(2x )2 − 6 × 2x − 16 = 0
Let t = 2x . So: t > 0.
t > 0, because for any value of x the power
2x is a positive number.
2
t − 6t − 16 = 0
∆ = 36 + 4 × 16 = 100
6 − 10
= −2
2
t1 =
t2 =
6 + 10
=8
2
2x = t2
t1 < 0
We look for positive solutions of the quadratic equation.
Number −2 does not meet the condition t > 0.
2x = 8, that is x = 3
PROBLEM
Solve the equation: 3 × 9x + 5 × 3x − 2 = 0.
Solving an exponential equation, it is not always possible to apply the
method shown in the examples above. Sometimes you can find a solution
using the logarithm’s definition.
EXAMPLE 4
Solve the equation.
a) 5 × 2x + 1 = 4
5 × 2x = 3
3
2x = 5
We use the definition of logarithm.
3
x = log2 5
b) 3x = 2 × 5x
3x
=2
5x
x
3
5
=2
We use the definition of logarithm.
x = log 3 2
5
PROBLEM
Solve the equation.
x
b) 3 × 5x = 10x
a) 4 × 3 = 7
Note that in example 4b) the equation could be transformed differently and
resultx = log 5 1 obtained.
3
2
EXPONENTIAL AND LOGARITHMIC EQUATIONS
MLR2-1 str. 93
93
By solving logarithmic equations, you can use the definition of logarithm
or logarithmic function’s injectiveness. When solving such equations, you
need to remember the appropriate assumptions.
Solve the equation.
EXAMPLE 5
a) 3 log5 (x + 1) + 4 = 6
Assumption x + 1 > 0
2
log5 (x + 1) =
3
We use the definition of logarithm.
2
x + 1 = 53
√
3
x = 25 − 1
We check the assumption:
b) log2 2x = log2 (x 2 − 3)
√
3
25 − 1 + 1 =
√
3
25 > 0.
Assumption 2x > 0 and x 2 − 3 > 0
2x = x 2 − 3
We use the condition: loga p = loga r
x 2 − 2x − 3 = 0
∆ = (−2)2 − 4 × 1 × (−3) = 16,
⇐
⇒ p = r.
√
∆=4
2+4
2−4
=3
= −1
x2 =
2
2
meets the assumption doesn’t meet the assumption
x1 =
PROBLEM
Solve the equation.
a) 5 log2 (x − 4) − 3 = 1
b) log 3x = log(x2 − 4)
If a > 0, b > 0, c > 0, k > 0,
a 6= 1 and k 6= 1, then:
loga (bc) = loga b + loga c
Sometimes, when solving logarithmic
equations, useful are the logarithm’s
properties, which we recall next to.
loga b = loga b − loga c
c
loga bp = p loga b
loga b =
Solve the equation.
EXAMPLE 6
1
1
1
a) log5 2 − 3 log5 x = log5 4 − 2
Assumption x > 0
1
1
1
log5 x = log5 2 + log5 52 − log5 4
3
1
log5 x = log5
3
1
2
× 25
1
4
logk b
logk a
2 = 2 log5 5 = log5 52
We use the rules of sum and difference
of logarithms.
log5 x = 3 log5 50
log5 x = log5 503
x = 125 000
94
Number 125 000 meets the assumption.
FUNCTIONS
MLR2-1 str. 94
b) 2 log2 x + log 1 x = 6
Assumption x > 0
4
2 log2 x +
2 log2 x +
log2 x
log2
1
4
=6
We use the theorem of changing the logarithm’s
base.
log2 x
=6
−2
3
log2 x = 6
2
log2 x = 4
We use the logarithm’s definition.
4
x =2
x = 16
Number 16 meets the assumption.
PROBLEM
Solve the equation.
b) log5 x + log√5 x = −6
a) log3 x − log3 4 = 2 log3 5 − 1
Note. By solving the equation of type ax = b, instead of using the logarithm’s definition, we can take the logarithm of both sides of such equation
(assuming any positive number other than 1 as the basis of the logarithm).
We say, ”we log each side”.
Below the equation 7x = 11 was solved in three ways. Each time the same
solution was received, only written in a different way.
7x = 11
7x = 11
7x = 11
log7 7x = log7 11
log11 7x = log11 11
log 7x = log 11
x log7 7 = log7 11
x log11 7 = 1
x log 7 = log 11
x = log7 11
x=
1
log11 7
x = log 11
log 7
Note. To calculate an approximate value of the solution of equation 7x = 11 using
a calculator, it is most convenient to use the decimal logarithm, i.e. use the last
method.
EXERCISE A
Write the solution of equation 5x = 3 using decimal logarithms.
EXERCISE B Two students were solving the equation 4x − 1 = 5x . Next
to, the beginning of both solutions
is presented. Find the results that
students received. Justify that the
numbers obtained are the same.
4
Student I
Student II
4x − 1 = 5x
4x − 1 = 5x
4x = 5x
4
4x = 4
5x
x
4
5
EXPONENTIAL AND LOGARITHMIC EQUATIONS
MLR2-1 str. 95
=4
log4 4x − 1 = log4 5x
(x − 1)log4 4 =x log4 5
x − 1 = x log4 5
x − x log4 5 = 1
95
PROBLEMS
1. Solve the equation.
x
1
5
= 125
4 x
b) 6 = 7
a)
7
e)
√ x
2 = 1
i) 2x ×
32
√ x √
f)
7 = 37
x
g) 1 = √1
6
x
c) 9 = 27
9
d) 1 = 8x
4
j)
√
2= 1
16
1 × 7x = √7
49
p
k) 0,2 = 25x
x
l) 4 = 8 × 3
3
h) 0,1 = 1000 × 10x
3
27
2. Solve the equation.
a) 7x = 3
d) 4 × 5x − 1 = 7
g) 23x − 1 = 5
b) 5 × 6x = 10
2x
e) 3 = 2
c) 3x + 1 = 4
f) 73x × 3 = 1
h) 42x + 3 = 3
x − 3
2
i)
=2
5
3
3. Solve the equation.
a) 2 log5 x = 6
c) 4 log6 x + 3 = 5
e) 1 − 2 log 1 x = 3
b) 3 + log 1 x = 1
d) 1 log4 x − 5 = −4
2
f) 7 − 3 log8 x = 9
2
3
3
5
4. Solve the equation.
a) log3 x = log3 5 + log3 4
e) 2 log 1 x = log 1 6 + log 1 24
b) log0,7 x = 3 log0,7 2 − log0,7 3
f) 4 log 2 = 2 log x + log 5
c) log5 4 + log5 x = 1 log5 49
g) log6 5 − 1 log6 x = log6 1
d) log 8 = log x − 2 log 1
h) log2 x2 − log2 x = 2 log2 5
2
2
2
2
5
2
2
5. Solve the equation.
a) logx + 1 36 = 2
b) log2 − x 1 = 3
8
8 = −3
125
d) log3x 15 = 1
2
e) log2 + x 5 = 1
2
√
3
f) log2x − 1 4 = − 1
b) log4 log3 (log2 x) = 0
c) log2 log2 (log2 x) = 1
b) xlog2 x = 64x
c) xlog3 x = 12
c) log x
2
3
6. Find x.
a) log2 (log3 x) = 2
7. Solve the equation.
a) xlog5 x = 25x
96
x
FUNCTIONS
MLR2-1 str. 96
APPLICATIONS OF EXPONENTIAL
AND LOGARITHMIC FUNCTIONS
By using exponential and logarithmic functions, phenomena from very
different fields of knowledge can be described.
SIZE OF POPULATION
A certain bacterial colony initially had 1000 bacteria, and their number
increased by 10 % every hour.
after the first hour: 1000 × 1,1
after the second hour: (1000 × 1,1) × 1,1 = 1000 × 1,12
after the third hour: (1000 × 1,12 ) × 1,1 = 1000 × 1,13
after t hours: 1000 × 1,1t
The colony of bacteria is an example of a population changing at a constant rate. Such changing populations can be described by formulas of the
form:
L(t) = b × at ,
where t is time, and constants a and b depend on the rate of population
change and its initial size. When we describe the population in this way,
we say that we have created its exponential model.
In 1971, 548 million people lived in India, and in 1991 — 846
million. The population of India in the years 1971–1995 changed according to
the exponential model.
EXAMPLE 1
a) Set a formula for this model.
L(t) = b × a t
L(t) — population of India (in mln)
t — time (in years) since 1971
548 = b × a 0
We write the general formula of exponential model.
The beginning of observation was 1971. Then t = 0 corresponds to 1971, i.e. L(0) = 548.
b = 548
846 = b × a 20
The year 1991 corresponds to t = 20, so L(20) = 846.
20
846 = 548 × a
r
846
a = 20
≈ 1,022
b = 548
548
We received the formula L(t) = 548 × 1,022t
APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
MLR2-1 str. 97
97
b) Estimate the population of India in 1980 and 1995.
L1980 = 548 × 1,0229 ≈ 667
L1995 = 548 × 1,02224 ≈ 924
We use the formula L(t) = 548 × 1,022t
for t = 9 and t = 24.
Ans. In 1980, the population of India was around 667 million, and in 1995 —
approx 924 million.
c) In 2000, the population of India exceeded 1 billion. Calculate in which year it
was to take place according to the given formula.
1000 = 548 × 1,022t
1000
1,022t = 548
1000
log 1,022t = log 548
We use the formula L(t) = 548 × 1,022t ,
for L(t) = 1000. 1 billion is 109 = 1000
million in both American and English numerical nomenclature.
We log both sides of the equation.
1000
t × log 1,022 = log 548
t=
log
1000
548
log 1,022
t ≈ 28
28 years after 1971.
Ans. According to the formula given, the population of India was to exceed
1 billion in 1999.
PROBLEM
The population of Nigeria in the years 2006–2017 changed according to
the exponential model. Set a formula for this model, knowing that 140 million people
lived in Nigeria in 2006 and 162 million in 2011. Using this formula, estimate Nigeria’s
population in 2015 and determine in which year the population will exceed 250 million.
Curiosity
In 1862, President of the United States Abraham Lincoln presented Congress
with a US population projection until 1930. On the basis of censuses from
1790 to1860, Lincoln noted that the natural growth rate was constant during
this period. So, he assumed that this ratio would not change until 1930, and
using an exponential model, he calculated that by then over 250 million
people would live in the US. In fact, the United States had 123 million
inhabitants in 1930.
As can be seen, the exponential model for calculating population growth is
effective only for short periods, because it assumes the invariance of the
birth rate. In fact, this ratio is not constant and depends on many factors,
including population size, cultural changes, migration of people, natural
disasters.
98
FUNCTIONS
MLR2-1 str. 98
The exponential model can also be used to describe other variable magnitudes than the population, but only those that change at a constant rate.
EXERCISE The mass M of a sample changes according to the exponential model. Determine whether this mass increases or decreases when:
a) M = 20 × 0,7t
b) M = 10 × 1,2t
LEVEL OF LOUDNESS
Some magnitudes are described from very small to very large numbers, so
that the range of values makes it difficult to use them.
An example is the sound intensity measured in W/m 2 . It was found that
the threshold of audibility, i.e. the lowest sound intensity that the average
person hears, is 10−12 W/m 2 . The loudest sound emitted on Earth was
accompanied by the eruption of the Krakatau volcano in Indonesia. It was
1023 W/m 2 .
The strength of a sound is determined by calculating how many times the
I intensity of this sound is greater than the I0 intensity of the sound corresponding to the hearing threshold. So, the sound strength is described by
the II0 quotient. From the information provided above, it follows that the
values of this quotient range from 1 to 1035 . A scale of this range would
be difficult to use. The decimal logarithms of these numbers take values
from 0 (log 1 = 0) do 35 (log 1035 = 35).
Numbers in such a reduced range are much easier to use, so a new concept
was introduced — the level of loudness. It is most often expressed in
decibels (abbreviated dB).
The loudness level (in decibels) can be calculated from the formula:
L(I) = 10 log I
I0
L(I) — level of loudness (in dB)
I — strength of sound in W/m2
I0 — strength of sound corresponding with the hearing threshold (I0 = 10−12 W/m2 )
The level of the hearing threshold is 0 dB and the volcanic eruption of
Krakatau had a sound level of 350 dB.
Note. The intensity of two simultaneous sounds is the sum of the intensities of
these sounds, but the loudness level of these two sounds is not the sum of their
loudness levels, because log (a + b) 6= log a + log b for positive a and b.
APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
MLR2-1 str. 99
99
The level of the kettle’s whistle is 90 dB and of the whistle of the
train is 110 dB.
EXAMPLE 2
a) How many times does the strength of the train’s whistle exceed the sound of
the kettle’s whistle?
I c — the strength
of the kettle’s whistle
90 = 10 log
9 = log
Ic
10−12
Ic
10−12
I p — the strength
of the train’s whistle
110 = 10 log
11 = log
Ip
10−12
Ip
10−12
Ic
= 109
10−12
Ip
= 1011
10−12
I c = 109 × 10−12
I p = 1011 × 10−12
I c = 10−3
h
W
m2
i
I p = 10−1
h
W
m2
We use the formula
L(I) = 10 log I ,
I0
where I 0 = 10−12 W/m2 .
We use the definition of logarithm.
i
Ip
10−1
=
= 100
Ic
10−3
Ans. The intensity of the train’s whistle is 10−1
the sound of the kettle’s whistle, which is 10−3
W
and is 100 times higher than
m2
W
.
m2
b) Calculate the loudness level of the whistles of two passing trains.
I — the volume of the whistles of two trains
I = I p + I p = 2 × 10−1
L = 10 log
2×10−1
=
10−12
The loudness levels are added. We have found
before that I p = 10−1 W/m2 .
We use the formula L(I) = 10 log I .
I0
= 10 log(2 × 10−1 − (−12) ) =
= 10 log(2 × 1011 ) =
log 2 × 1011 = log 2 + log 1011
= 10(log 2 + log 1011 ) =
= 10(log 2 + 11 log 10) =
= 10(log 2 + 11) ≈ 113
Ans. The sound level of two trains’ whistles is around 113 dB.
100
FUNCTIONS
MLR2-1 str. 100
c) How many kettle whistles create a painful noise in the ear, i.e. 130 dB?
I n — sound intensity of n kettles whistling
I n = n × I c = n × 10−3
130 = 10 log
We have found that I c = 10−3 W/m2 .
n ×10−3
10−12
We use the formula L(I) = 10 log I .
I0
13 = log(n × 109 )
We use the definition of logarithm.
n × 109 = 1013
n=
1013
109
n = 104
Ans. It takes as many as 10 000 whistling kettles to bring the volume up 130 dB.
PROBLEM
The sound produced by a refrigerator is equal to 40 dB, and by a vacuum cleaner to 80 dB.
a) How many times does the loudness produced by the vacuum cleaner exceed the
sound produced by the refrigerator?
b) Calculate the loudness of a refrigerator and vacuum cleaner working simultaneously.
c) How many working vacuum cleaners create noise at 100 dB?
PROBLEMS
1. The
population of Poland in the years 1962–1970 changed according to the
exponential model. Use the data in the table and compose the formula for this
model.
Year
1962
1964
Population of Poland
30,5 mln
31,3 mln
a) Estimate what was (according to the formula) the population of Poland in 1970.
Compare your result with the actual population this year (32,6 million).
b) Estimate what the number of Polish inhabitants would be in the current year
if the natural growth in the period 1962–1964 remained unchanged until today.
Compare the result obtained with the actual population of Poland.
c) In which year, according to this model, would the population of Poland exceed
50 million if the birth rate of 1962–1964 remained unchanged?
APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
MLR2-1 str. 101
101
2. A
patient has taken a 50 mg dose of a certain drug. It is known that within 6
hours 60 % of this drug is removed from the bloodstream.
a) The mass m of the drug (in mg) remaining in the body after time t (in hours) can
be calculated from the formula: m = abt , where a and b are constants. Calculate
the values of these constants.
b) How much medicine remains in the bloodstream after an hour, and how much
— after a day?
c) The patient should take a second dose before the first dose in the bloodstream
falls below 10 mg. How many hours after taking the first dose should he take the
second dose?
3. The next box contains the loudness levels
rustling leaves — 10 dB
of several selected sounds.
a) Calculate the sound intensity of rustling
leaves.
b) How many times less intense is the sound
of pianissimo playing the violin than the
sound of a fortissimo playing orchestra?
violin (pianissimo) — 30 dB
scream — 80 dB
pneumatic hammer — 100 dB
orchestra (fortissimo) —100 dB
rock concert — 120 dB
c) Calculate the sound level of two pneumatic hammers working simultaneously.
d) Calculate the sound level of the concert performed simultaneously by nine violinists playing pianissimo.
e) Calculate the noise level to which the person sitting next to the screaming spectator at a loud rock concert is exposed.
f) Calculate how many working pneumatic hammers produce a noise equal to the
volume of a rock concert.
Curiosity
An object placed in an environment at a lower temperature than the object’s temperature will begin to cool down. When the ambient temperature is constant, the
temperature T of the object after time t is described by the formula:
T (t) = T0 + (Tp − T0 )at
T0 — environment’s temperature (in ◦C)
Tp — initial temperature of the object (in ◦C)
a — the constant characteristic for the object
4. The formula
T = T0 + (Tp − T0 )at as presented in the curiosity is used by forensic medicine specialists, specifying the time that has elapsed since the death of
the deceased. Let’s assume that the police found the body at 1800 , the victim’s
body temperature was 30◦, and the ambient temperature, like the whole afternoon,
was 10◦C. Let’s also assume that after an hour the ambient temperature has not
changed, but the body has cooled to 28◦C. Calculate what time he died.
102
FUNCTIONS
MLR2-1 str. 102
Figures
on the plane. Part 2
Imagine that you are standing on the beach on a sunny day looking at the see.
How far from your eyes is the horizon?
Area of a disc. Length of a circle
and inscribed angles
about a polygon
Properties of central
Lines and circles
Circle inscribed in a polygon
Properties of polygons. Regular polygons
MLR2-1 str. 103
Circle circumscribed
AREA OF A DISC. LENGTH OF A CIRCLE
EXERCISE A How to draw a circle on the sand with a string at your disposal?
Describe in words what figure we call a circle and which figure — a disc. What
is the chord of a circle and what is the diameter?
Let us remind you that a circle with center S and radius r is the set of
points of the plane whose distance from the point S is equal to r .
A disc with center O and radius r is the set of points on the plane whose
distance from point O is less than or equal to r.
EXERCISE B Let c(S, r ) mean the circle with center S and radius r and let
d(S, r ) mean the disc with center S and radius r . Points belonging to the circle
meet the condition: P ∈ c(S, r ) ⇐
⇒ |P S| = r . Write a similar condition for points
on the disc.
Worth knowing!
The center of a circle lies on the perpendicular bisector of each chord of the
circle, as it is equally distant from the
ends of this chord. Therefore, to designate
the center of a given circle, it is enough
to draw two non-parallel chords and construct their perpendicular bisectors. The
intersection of these lines is the sought
center of the circle.
Already in ancient times it was noticed
that the ratio of the length of a circle to
the length of its diameter is the same
for all circles (regardless of the size of
the circle). A number equal to this ratio
is irrational and we denote it with the
letter π .
circle’s length
diameter’s length
Length of circle:
l = 2πr
=π
Here are the first several digits of the
decimal expansion of this number:
π = 3,14159265358979 . . .
Area of disc:
A = πr 2
104
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 104
From history
Letter π as the name of the ratio of the length of circle to the length of its
diameter maybe came from the fact that it is the first letter of the Greek
′
word περιµ
ετρoς (perimetros) — perimeter.
Various approximations of the π number have been used for centuries.
E.g, from a passage in the Bible (I Kings, chapters 7, 23) it appears that in
biblical times, π ≈ 3 was accepted. Egyptians (20th century B.C.) accepted
2
22
16
π ≈ 9 , and Archimedes (3rd century B.C.) adopted π ≈ 7 .
Also, in modern times, attempts have been made to provide the most accurate value of π. In 1610 Dutch scholar Ludolph van Ceulen gave 35 digits
after the decimal point. In his honor the number π is sometimes called
ludolfine.
Using computers, around 22,5 billion digits after the decimal point of π
were determined in 2016. The calculations took 105 days. Number π is
irrational, which means that the digits of its decimal expansion appear
irregularly. An interesting fact is that in this expansion you can find any
sequence of several digits. For example, the sequence 01052004, i.e. the
subsequent digits of the date of Poland’s accession to the European Union,
appear in the first 100 million digits after the decimal point. You can check
this by using the appropriate computer program available on the internet.
The formula that allows you to calculate the length l of a circle results
directly from the definition of number π . (Because π = l , so l = 2π r ).
2r
We will now show how one could try to justify a disc’s area formula.
Imagine that a circle with a radius r we divide into equal parts (see figure). Each area of these parts
is not much different from the triangle’s area with
basis of a and height r , i.e. is approximately equal
a×r.
2
If the number of parts into which we divide the
circle is equal to n, then the base of the triangle is
approximately equal to 1 of the length of the circle,
n
i.e.
1 × 2π r = 2π r
a≈ n
n
The area of each part is 1 the area of the disc. Thus:
n
area of disc ≈ n × a × r ≈ n ×
2
2π r
n
×r
2
= πr2
The more parts we divide the disc into, the approximation will be the more accurate. We can therefore
assume that the disc’s area is equal to π r 2 .
AREA OF A DISC. LENGTH OF A CIRCLE
MLR2-1 str. 105
105
EXERCISE C Look at the picture. How many times
the length of the arc of part marked in blue is less
than the length of the whole circle if α = 108◦?
What part of the area of the whole disc is the area
of the marked figure?
The angle with the vertex in the center of the circle is called a central
angle. The common part of the central angle and the disc is a sector, and
the common part of this angle and the circle is the circular arc. We say
that the central angle is based on this arc.
An arc’s length is a fraction of the length of the circle, and the area of a disc’s
sector is a fraction of the disc’s area. This fraction is equal to α ◦.
360
Length of circular arc:
l =
α × 2πr
360◦
Area of sector:
A=
α × πr 2
360◦
EXAMPLE
a) The central angle in a circle with the radius of 9 has a measure
of 140◦. Calculate the area of the sector determined by this angle.
Area of the sector =
140◦
7
63
× π × 92 = 18 × 81 × π = 2 π
360◦
b) In the figure a sector is shown. Calculate the perimeter of this figure.
α = 360◦ − 160◦ = 200◦
The arc is cut out by the central angle
α with measure 200◦.
200◦
5
10
× 2π × 3 = 9 × 6π = 3 π
360◦
10
The perimeter of the sector is = 6 +
π
3
The length of arc =
PROBLEM
A central angle of 40◦ is marked in a circle with the radius of 10.
Calculate the area and perimeter of the sector determined by this angle.
106
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 106
PROBLEMS
1. Calculate:
a) area of a disc with a radius of 7,
d) radius of a disc with area 10,
b) area of a disc with a diameter of 12,
e) radius of a circle of length 5,
c) length of a circle with a diameter of 17,
f) circle length with inside area 4.
2. Write the formula that allows you to calculate:
a) the length l of the circle, when its diameter d is given,
b) the area A of the disc, when its length l is given,
c) the diameter d of the disc when its area A is given,
d) circle’s length l when its inside area A is given.
3. The minute hand on a certain clock is 8 cm long. Calculate the path that the end
of this tip takes over:
a) an hour,
b) a day,
c) a quarter of an hour,
d) 45 minutes.
4. Calculate the areas of the shaded figures. Assume that the side of the grille is 1.
5. a) Write down the length of the side of the square which has the same area as
the disc with radius r .
b) A disc and a square have equal areas. Which figure has larger perimeter? How
many times?
Curiosity
How to construct a square with an area equal to the area of a given disc using
a compass and a ruler? This problem, called quadrature of the circle, has been
tried to be solved for over 2000 years. Only in the nineteenth century it was
proved that such a construction is impossible. Thus, it was proved that for a given
√
line segment r it is impossible to construct a line segment r π long. In many
languages, ”quadrature of the circle” figuratively means an unsolvable problem.
AREA OF A DISC. LENGTH OF A CIRCLE
MLR2-1 str. 107
107
6. Calculate the lengths of the selected arcs and the areas of the shaded figures.
7. The pendulum of the old clock is 50 cm long and
deviates from the vertical by 18◦. Full swing (from
left to right and back) takes 2 sec. What path does
the end of the pendulum cover during an hour?
8. Angle
measures can be expressed in units other
than degrees, such as radians. The measure of the
center angle is 1 radian when the arc on which this
angle is based has the same length as the circle’s
radius.
a) 1 radian — how many degrees is it?
b) 90◦ — how many radians?
PROPERTIES OF CENTRAL ANGLES
AND INSCRIBED ANGLES
The drawing beside indicates the central angle α and the arc on which this angle is
based.
In this drawing you can indicate one more central
angle, its measure is 360◦ − α. The arc on which
this angle is based is marked in black.
The inscribed angle is the angle whose vertex lies
on the circle, the arms intersect the circle, and the
measure is less than 180◦. The adjacent figure indicates the inscribed angle β and the arc on which
this angle is based.
Note that there is only one central angle based on
a given arc, while there are infinitely many inscribed
angles based on that arc.
108
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 108
Theorem on inscribed and central angles
based on the same arc
The inscribed angle is twice as small
as the central angle based on the same arc.
Proof
Let’s assume that angle α is the inscribed angle and angle β is the central
angle based on the same arc as the angle α. We will show that β = 2α.
There are three possible cases of positioning the center of the circle relative to
the angle α: the center of the circle may lie on the arm of α, it may lie inside
this angle, it may also lie outside the angle.
1. Let the center of the circle lie on the arm
α, as shown in the figure. The triangle BCO is
isosceles, so:
| ¾ BOC| = 180◦ − 2α
The angle β and ¾ BOC are supplementary angles, so:
β = 180◦ − | ¾ BOC| = 180◦ − (180◦ − 2α) = 2α
2. Let the center of the circle lie within α, as shown in the first figure below.
Drawing the diameter from the vertex
of angle α, we divide α into two angles
γ and δ. Then the angle β will also be
divided into two angles. From previous
considerations we know that these angles are 2γ and 2δ, so:
α=γ+δ
β = 2γ + 2δ
Hence:
β = 2(γ + δ) = 2α
3. Let the center of the circle lie outside the inscribed angle α, as shown in the
first figure below.
Drawing the diameter from the vertex of angle α, we get the situation
presented in the second drawing. The
inscribed angle α + γ is based on the
same arc as the central angle β + 2γ.
From previous considerations we know
that:
β + 2γ = 2(α + γ)
Hence:
β = 2(α + γ) − 2γ = 2α
In each of the considered cases we received β = 2α.
PROPERTIES OF CENTRAL ANGLES AND INSCRIBED ANGLES
MLR2-1 str. 109
109
Theorem on inscribed angles
based on the same arc
Inscribed angles based on the same arc
have equal measures.
Proof
From the previous theorem it follows that when the inscribed angles are based on
the same arc, each of them has a measure two times smaller than the central
angle based on this arc. So, all these inscribed angles have equal measures.
If the arcs of a given circle have equal lengths, then
the central angles that cut them out have equal measures. It follows that the inscribed angles are also
equal if they are based on arcs of the same length.
When the central angle or inscribed angle is based on
a certain arc, it can also be said that it is based on the
chord connecting the ends of the arc.
Theorem on inscribed angle
based on diameter
The inscribed angle based on the diameter
is a right angle.
Proof
The central angle based on the semi-circle is 180◦. The inscribed angle based on
the diameter has a measure two times smaller than it, i.e. it is a right angle.
PROBLEMS
1. Provide angle measures α, β, γ
110
and δ.
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 110
2. Provide angle measures α, β, γ
and δ.
3. Look at the picture. Justify that the sum of the measures of inscribed angles α
and β based on the same chord is equal to 180◦.
4. Points
A, B, C, D lie on a circle and |AB| = |CD|. Prove that the quadrilateral
whose vertices are these points is a trapezoid.
5. Calculate the angles of the polygon, whose sides are marked in blue.
6. Look
at the figure below. The acute
angles of the setsquare have measures
30◦ and 60◦. What part of the circle
does that setsquare cover?
7. Points O and S in the figure below
are the centers of the circles drawn.
What is the angle α?
PROPERTIES OF CENTRAL ANGLES AND INSCRIBED ANGLES
MLR2-1 str. 111
111
LINES AND CIRCLES
A line and a circle may not have common points, they may intersect at
two points, or they may have one common point. A line that has only one
point in common with a circle is called a tangent to the circle.
Theorem
The tangent to the circle is
perpendicular to the radius
at the point of tangency.
Proof
The proof will be carried out using the indirect method.
Suppose one of the angles between the tangent to the circle and the radius to the point of contact is acute. In the
drawing, it is the angle α with vertex A. Drawing from the
center S of the circle another line, inclined to the tangent
at angle α, we get the isosceles triangle ABS.
Then |SB| = |SA| and point B would have to belong to the circle, so it would
be the second common point of the tangent and the circle, which contradicts
the definition of the tangent.
Theorem
When the tangents to the circle
intersect, the line segments connecting
the intersection point with the
tangency points are equal in length.
Proof
Let the lines P A and P B be tangent to the circle
with center S at points A and B. Then the triangles P SA and P SB are rectangular, they have a
common hypotenuse P S and |SA| = |SB|. According to Pythagoras’ Theorem, |P A| = |P B|.
112
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 112
Theorem on the angle between tangent and chord
The acute angle α between the chord of the circle
and the tangent at the endpoint of the chord is
equal to the inscribed angle based on this chord.
α=β
Proof
Let us assume that the line m and the circle with
center S are tangent at point A and α is the
acute angle between line m and chord AB (see
the drawing on the right).
Then:
|¾ SAB| = 90◦ − α
The SAB triangle is isosceles, so:
|¾ ASB| = 180◦ − 2 × |¾ SAB| = 180◦ − 2(90◦ − α) = 2α
Angle β is the inscribed angle based on the chord AB, so:
β = 1 |¾ ASB| = 1 × 2α = α
2
2
Good to know!
Let O be the center of the circle. We want to draw a tangent to this circle
through some point P .
When point P lies on the circle, it
is sufficient to draw the ray OP and
construct a line perpendicular to it
passing through point P (see first figure).
When point P lies outside the circle,
we first determine the center of the
segment OP and draw a circle whose
diameter is OP . Then we draw a line
through P and the common point of
the circles (see the second figure).
The line drawn is perpendicular to the
radius of the given circle, because it is
known that the inscribed angle based
on the diameter OP is a right angle.
So, this line is the tangent sought.
Of course, by drawing a line through P and the second common point of
the circles, we also get a tangent to the circle.
LINES AND CIRCLES
MLR2-1 str. 113
113
Two different circles can be placed relative to each other so that they have
no common point or have two common points or only one common point.
When circles have only one point in common, we say they are tangent.
Note. Circles that have a common center are said to be concentric. If the concentric circles have different radii, then these circles are of course disjoint. If the
radii of the concentric circles are the same, then the circles have infinitely many
common points (coincide).
EXERCISE A Draw a circle with a radius of 2 cm. Label its center with the letter
O and draw through O any line. Using compasses, draw a new circle with
center on the drawn line, tangent to the drawn circle.
A line passing through the centers of two tangent circles also passes
through their point of contact. It follows that the distance between the
centers of the tangent circles is equal to the sum or difference of the
lengths of radii of these circles.
EXERCISE B A circle with center S has a radius of 2 and a circle with center T
has a radius of 5. Determine the mutual position of these circles if:
a) |ST | = 7
114
b) |ST | = 8
c) |ST | = 4
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 114
EXERCISE C How many lines can be tangent to two circles at the same time?
Consider the different positions of the circles.
Curiosity
In the figure below, the circles are externally disjoint, and each of the four
straight lines is tangent to both circles. We will show you how to construct
such lines.
Let the circle with center S and radius R lie outside the circle with center O
and radius r . Let us assume that R > r .
The method of constructing line a: We draw a circle with the center S and
the radius R − r , then we draw from point O a tangent line to this circle (as
described on p. 113). Through the obtained tangent point A we draw a ray
SA, which intersects the large circle at point B. Drawing through point B
a line perpendicular to the ray SA, we get the tangent sought, because the
AOCB is a rectangle.
The method of constructing line c: We draw a circle with the center S and
the radius R + r , then we draw a tangent to this circle passing through
the point O. We draw radius SK (where K is the tangent point). Through
the point L — the common point of the circle with the radius R and SK
line segment — we run a line perpendicular to SK. This line is the tangent
sought.
Note that line b can be constructed like line a, and line d can be constructed
like line c.
LINES AND CIRCLES
MLR2-1 str. 115
115
PROBLEMS
1. The lines in the figure below are parallel by pairs. Distances between neighboring lines are equal to 1.
a) A circle with center A has exactly three common points with lines drawn. What is the radius
of this circle?
b) Can a circle with center B have exactly three
points in common with these lines?
c) What radius can a circle with center C have to
be cut by exactly one line?
d) What radius should a circle with center D have
so that it is crossed by four lines (each at two
points)?
e) What radius can a circle with center E have if it is known that the circle has
exactly four points in common with drawn lines?
f) How many lines cross the circle with center P and radius 2?
2. Line m is tangent to the circle. Determine the measure of angles marked with
an arc.
3. The
arms of the angle β in the adjacent
figure are tangent to the circle. What is the relationship between angle measures α and β?
4. The sides of the triangle drawn are tangent to the circle. Calculate the perimeter
of this triangle.
116
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 116
5. The drawn line is tangent to the circle. Express the measure of angle β depending on the measure of angle α.
6. The
circles in the figure are tangent by pairs. The largest of the circles
has a radius of length r . Calculate the
perimeter of the OP S triangle.
7. The circles in the figure below are
tangent by pairs. The radius of the
largest of them has the length of r .
Calculate the length of the radius of
the smallest of these circles.
CIRCLE CICUMSCRIBED ABOUT A POLYGON
We say that the circle is circumscribed
about polygon when all the vertices of
that polygon are on the circle.
If the circle is circumscribed about polygon, we can also say that the polygon is
inscribed in the circle.
The center of the circle circumscribed about a polygon is a point equidistant from its vertices. It follows from the property of perpendicular
bisector (see p. 54) that this center must lie on the perpendicular bisector of each side of the polygon.
CIRCLE CICUMSCRIBED ABOUT A POLYGON
MLR2-1 str. 117
117
We already know that the perpendicular bisectors of sides of each triangle
intersect at one point (see p. 54). The following theorem follows:
Twierdzenie
A circle can be circumscribed about
each triangle. The center of such a
circle is the intersection point of the
perpendicular bisectors of sides of the triangle.
To constructively identify the center of the
circumscribed circle of a triangle, it is sufficient to draw the perpendicular bisectors of
two sides of the triangle. Their intersection is
the center of the sought circle, and the distance of this point from any vertex of the
triangle is the radius of the circle.
EXERCISE A Draw a triangle and construct the
circumscribed circle of this triangle.
Note that the properties of the central and
inscribed angles show that the center of the
circumscribed circle of the right-angled triangle is the center of the hypotenuse.
EXERCISE B a) Calculate the radius of the circle circumscribed about the rightangled triangle of 4 and 5 legs’ length.
b) Calculate the sides’ length of the isosceles right-angled triangle inscribed in
a circle with a radius of 20.
You can’t circumscribe a circle around every polygon. It is possible only
when perpendicular bisectors of all its sides intersect at one point. For
example, a circle cannot be circumscribed about a parallelogram that is
not a rectangle.
EXERCISE C About which of the drawn polygons can you certainly not circumscribe a circle?
118
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 118
Theorem
A circle can be circumscribed about a quadrilateral
if and only if
the sums of measures of opposite angles are equal.
α + γ = β + δ = 180◦
Proof
We will first demonstrate that if a circle can be circumscribed about a quadrilateral, then the sum of measures of the opposite angles of the quadrilateral
is 180◦.
Let α and β denote the opposite angles of the
quadrilateral inscribed in the circle (see figure).
The central angles based on the same arcs as the
angles α and β respectively have measures 2α
and 2β and 2α + 2β = 360◦, hence α + β = 180◦.
The theorem on the sum of measures of quadrilateral’s angles says that the sum of the other two
angles of the quadrilateral is also 180◦.
We will now show that if the sum of measures of opposite angles in a quadrilateral is 180◦, then about this quadrilateral a circle can be circumscribed.
Let us assume that in the quadrilateral ABCD:
|¾ BAD| = α, |¾ BCD| = β and α + β = 180◦
(see figure).
Note that all angles of the ABCD quadrilateral
are convex, so point C lies within the angle α.
Consider the circle circumscribed about triangle ABD. Suppose that the distance of point C
from the center of this circle is greater than the
radius of the circle.
Let P be a point on the arc of the circle on which angle α is based, and
P 6= B and P 6= D. Then the inscribed angle BP D has the measure 180◦ − α and
therefore in the quadrilateral BCDP the sum of the angles with vertices P and
C would be:
|¾ BP D| + |¾ BCD| = 360◦ − (180◦ − α) + β = 180◦ + α + β = 180◦ + 180◦ = 360◦
Such a quadrilateral cannot exist, so the distance of point C from the center
of the circumscribed circle about triangle ABD is not greater than the radius
of this circle.
Similarly, it can be shown that the distance of point C from the center of
the considered circle cannot be smaller than the radius of the circle. Point
C must therefore lie on this circle, i.e. it is a circumscribed circle about the
quadrilateral ABCD.
CIRCLE CICUMSCRIBED ABOUT A POLYGON
MLR2-1 str. 119
119
EXAMPLE
The consecutive angles of a pentagon inscribed in a circle have
the following measures: 100◦, 120◦, 110◦, 100◦ and 110◦. In this pentagon there
are diagonals coming out from the vertex of 120◦. To what angles do these
diagonals divide this angle?
We draw an auxiliary drawing.
Quadrilateral ABCD is inscribed in a circle, so
|¾ ABD| + |¾ AED| = 180◦.
α + β + 110◦ = 180◦
α + β = 70◦
γ = 120◦ − (α + β) = 120◦ − 70◦ = 50◦
|¾ ABC | = α + β + γ = 120◦
β + γ + 100◦ = 180◦
Quadrilateral BCDE is inscribed in a circle, so
|¾ CBE | + |¾ CDE | = 180◦.
β + γ = 80◦
β = 80◦ − γ = 80◦ − 50◦ = 30◦
α = 120◦ − (β + γ) = 120◦ − 80◦ = 40◦
Ans. The diagonals divide the 120◦ angle into 40◦, 30◦ and 50◦ angles.
PROBLEM
The measures of angles of ABCDE pentagon inscribed in a circle are:|¾ ABC| = 110◦, |¾ BCD| =
95◦, |¾ CDE| = 85◦, |¾ DEA| = 120◦, |¾ EAB| = 130◦. Find
BEC angle measure.
PROBLEMS
1. Determine if the center of the circumscribed circle about a triangle belongs to
the triangle when measures of two angles of the triangle are:
a) 39◦ and 47◦
b) 70◦ and 55◦
c) 63◦ and 27◦
2. a) Calculate the length of the radius of the circumscribed circle about the rightangled triangle with catheti lengths 5 and 12.
b) In the right-angled triangle, the legs are 6 and 8 long. Calculate the length of
the line segment connecting the vertex of the right angle with the center of the
hypotenuse.
120
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 120
3. Which of the following statements are true?
1 The center of the circle circumscribed about the triangle always lies at the intersection of the lines including heights of this triangle.
2 The center of the circle circumscribed about an equilateral triangle is the point
of intersection of the heights of this triangle.
3 For any three non-collinear points, you can draw a circle that passes through
these points.
4 Each side of the acute-angled triangle inscribed in a circle is shorter than the
diameter of that circle.
5 Each side of the obtuse-angled triangle inscribed in a circle is shorter than the
diameter of that circle.
4. a)
Calculate the radius of the circle circumscribed about the rectangle with sides
5 cm and 12 cm.
b) Calculate the side length of the square inscribed in a circle with radius 5 cm.
5. Calculate the angle measures of the quadrilateral drawn.
6. The picture shows the triangles ABC and ABD. Justify that the circumscribed
circle about the ABC triangle has the same radius as the circle circumscribed about
the ABD triangle.
7. Angle
measures of a quadrilateral expressed in degrees are consecutive odd
numbers. Can you describe a circle about this quadrilateral?
8. Point
P is symmetrical to the orthocenter of triangle ABC relative to line AB.
Prove that point P lies on the circle circumscribed about the triangle ABC.
9. Prove that if the bisectors of all quadrilateral’s angles intersect at four points,
then these points form a quadrilateral about which you can circumscribe a circle.
CIRCLE CICUMSCRIBED ABOUT A POLYGON
MLR2-1 str. 121
121
CIRCLE INSCRIBED IN A POLYGON
We say that a circle is inscribed in a
polygon, if it is tangent to every side
of this polygon.
If the circle is inscribed in a polygon, we
can also say that the polygon is circumscribed about the circle.
The distance of the center of the circle inscribed in a polygon from each
side of the polygon is equal to the radius of the circle. The center of such a
circle is equally distant from all its sides, so it follows from angle’s bisector
property (see p. 55) that this center must lie on the bisector of each angle
of the polygon.
Since we already know that the bisectors of the angles of a triangle intersect at one point, the following statement is true:
Theorem
In each triangle a circle can be inscribed. The
center of the circle inscribed in a triangle is the
intersection point of the bisectors of the triangle’s angles.
To construct a circle inscribed in a triangle, we
draw the bisectors of any two angles of the triangle, their common point S will be the center
of the searched circle. Then we draw a line passing through the S point and perpendicular to
one of the sides. The point of intersection of
this line with the AC side (K in the drawing) is
one of the tangent points. We draw the circle
with the center S and radius |SK|.
EXERCISE A
Draw a triangle and construct a circle inscribed in this triangle.
122
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 122
You can’t inscribe a circle in every polygon. This is only possible if the
bisectors of all angles intersect at one point. For example, you cannot
inscribe a circle in a rectangle that is not a square.
EXERCISE B
circle?
In which of the drawn polygons can you definitely not inscribe a
The condition that a quadrilateral circumscribed about a circle must meet
is as follows:
Theorem
A circle can be inscribed in a
quadrilateral if and only if the
quadrilateral is convex and the sums of
lengths of its opposite sides are equal.
Proof
Suppose you can inscribe a circle in a quadrilateral. This quadrilateral must be
convex because it cannot have angles greater than 180◦. We will show that the
sums of lengths of its opposite sides are equal.
The sides of the quadrilateral are tangent to the
circle. From properties of a tangent to a circle
(see p. 112) it follows that the line segments
connecting vertices to tangent points have equal
lengths (see figure).
It is easy to see that regardless of which pair of
opposite sides of the quadrangle we choose, the
sum of their lengths is x + y + z + t, i.e. the sums
of the lengths of the opposite sides are equal.
Let now ABCD be a convex quadrilateral in which the sums of the lengths of
the opposing sides are equal, i.e. |AB| + |CD| = |BC| + |AD|. We will show, by
the indirect method, that a circle can be inscribed in this quadrilateral.
Let S be the intersection point of the bisectors of BAD and ABC angles; let
also the circle with center S be tangent to sides AB, BC and AD (see figure).
Suppose this circle is not tangent to the side CD.
CIRCLE INSCRIBED IN A POLYGON
MLR2-1 str. 123
123
The tangent to the circle drawn through
point C will cut line AD at point D′ different
from D. We can assume that |AD′ | > |AD|
(if |AD′ | < |AD|, the proof would be analogous), i.e. points C, D, D′ should be the
vertices of a triangle.
The quadrilateral ABCD′ is circumscribed on
the circle, so:
|AD′ | + |BC| = |AB| + |CD′ |
We assumed that the following condition was also met:
|AD| + |BC| = |AB| + |CD|
Both conditions result in equality:
|AD′ | − |AD| = |CD′ | − |CD|
|AD′ | − |AD| = |DD′ |, so |DD′ | = |CD′ | − |CD|.
Hence |DD′ | + |CD| = |CD′ |.
Thus, the line segments CD, DD′ , D′ C do not satisfy the triangle inequality,
i.e. there is no triangle with vertices C, D, D′ (contradiction). Therefore D = D′ .
So, the ABCD quadrilateral is circumscribed about the circle.
We will now show how one can calculate the area of a polygon circumscribed about a circle, when we know the radius of this circle and the
perimeter of the polygon.
In the drawing, the pentagon circumscribed about
the circle is divided into triangles by line segments
connecting the vertices with the center of the circle.
The area of the polygon is equal to the sum of the
areas of these triangles:
A = 1 ar + 1 br + 1 cr + 1 dr + 1
er
2
2
2
2
2
A = 1 r (a + b + c + d + e)
2
We conducted the above reasoning for a pentagon. A similar reasoning
could be made for any polygon. Therefore:
Theorem
The area of a polygon
circumscribed about a circle
is equal to the product of the
radius of this circle by half
the perimeter of the polygon.
124
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 124
EXAMPLE
The AC diagonal of a quadrilateral ABCD is 6 long and divides the
quadrilateral into an equilateral triangle and an isosceles right-angled triangle
with the hypotenuse AC . You can inscribe a circle in this quadrilateral. What
radius has such a circle?
|AD| = |DC | = 6
√
6
|AB| = |BC | = √ = 3 2
2
AABCD = AABC + AACD =
√
p =6+3 2
r=
√
√
AABCD
3 + 3√ 3
9 + 9√3
=
=
p
6+3 2
2+ 2
√
√
√
√
3 2×3 2
62 3
+
= 9+9 3
2
4
p — half the perimeter of ABCD quadrilateral
r — radius of the inscribed circle in ABCD
quadrilateral
PROBLEM
In the KLMN quadrilateral, the diagonal KM is 24 long and divides the
quadrilateral into two isosceles triangles with base KM. The arms of these triangles are
13 and 20 long. What is the radius of the circle inscribed in the KLMN quadrilateral?
PROBLEMS
1. The following drawings are triangles: acute-angled, isosceles and right-angled.
Find the measures of angles of these triangles and the measures of the angles
labelled with letters.
2. Draw any acute angle. Mark point A on one of the arms. Construct a circle
tangent to both arms of the angle drawn so as point A be one of the tangency
points.
3. In the isosceles triangle ABC, in which the angle between the arms AC and BC
is 100◦, a circle with the center O is inscribed. Calculate the measure of the angle
AOB.
CIRCLE INSCRIBED IN A POLYGON
MLR2-1 str. 125
125
angle between the arms AC and BC of the isosceles triangle ABC is 40◦
wide. Point O is the center of the circle inscribed in triangle ABC, and point S is
the center of the circle circumscribed about this triangle. Calculate |¾ SAO|.
4. The
5. In
the figure beside, the triangle with angles α
and β is circumscribed on the circle. The angle γ
is an angle of the triangle whose vertices are the
points of tangency. Show that:
γ = α+β
2
6. Calculate the perimeter of the quadrilateral shown in the figure.
7. Points
A, B and C are vertices of triangle ABC where |AB| = |BC|. Construct
such a point D that there is a circle inscribed in the quadrilateral ABCD and there
is a circle circumscribed about this quadrilateral.
8. Justify that the area of a quadrilateral circumscribed on the circle with radius r
is r (a + b), where a and b are the lengths of two opposite sides.
PROPERTIES OF POLYGONS.
REGULAR POLYGONS
EXERCISE A
Sketch:
a) a quadrilateral in which two non-adjacent sides are perpendicular,
b) a pentagon which has one pair of parallel sides,
c) a hexagon in which sides are parallel by pairs.
Note that some polygons satisfy the
condition: for any two points of the
polygon, the line segment whose ends
are these points is included in this
polygon. These polygons are called
convex polygons.
126
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 126
EXERCISE B
Here are examples of different polygons. Which of them are convex?
Note that all angles of the convex polygon are angles greater than 0◦ and
less than 180◦ (these are convex angles). In a non-convex polygon, at least
one angle is a non-convex angle.
Let us remind you that the diagonal of a polygon is each line segment
connecting the non-neighboring vertices of the polygon. Some diagonals
of a non-convex polygon lie outside
the polygon.
We already know that the sum of triangle angles’ measures is 180◦, and
the sum of the quadrilateral angles’ measures is 360◦. Let’s consider what
we can say about the sum of measures of the angles of an n-gon, i.e. a
polygon that has n vertices.
EXERCISE C Draw any convex polygon. Choose any vertex and draw all the
diagonals coming out of it. How many triangles the polygon was divided into
by the drawn diagonals?
EXERCISE D Look at the picture. Into how many
triangles the diagonals divided this 11-gon? What
is the sum of 11-gon angles’ measures?
All diagonals drawn from one vertex of a convex n-gon divide this polygon
into n − 2 triangles (there are as many triangles as sides that do not contain
the selected vertex). The sum of the angle measures of these triangles is
(n − 2) × 180◦ and is equal to the sum of the angle measures of the n-gon.
The same relationship applies to non-convex polygons.
Theorem
The sum of the angle measures of an n-gon is (n − 2)×180◦.
PROPERTIES OF POLYGONS. REGULAR POLYGONS
MLR2-1 str. 127
127
EXERCISE E
Draw any hexagon and three diagonals from each vertex. If for
a certain vertex a diagonal has already been drawn, draw it again, but with a
different color. How many times were each of the diagonals drawn? How many
diagonals does the hexagon have?
You can draw n − 3 diagonals from each vertex of the n-gon. (You cannot
draw the diagonal to it or to the adjacent vertices from the selected vertex.)
One can imagine that by drawing the diagonals from each vertex, we would
draw n(n − 3) segments, each twice. This reasoning leads to the following
theorem:
Theorem
The number of diagonals in n-gon is n(n − 3) .
2
EXAMPLE
How many diagonals has a polygon in which the sum of the angle
measures is 2700◦?
(n − 2) × 180◦ = 2700◦
We use the theorem on the sum of angle measures of n-gon.
n − 2 = 15
The considered n-gon is a 17-gon.
n = 17
number of diagonals =
17(17 − 3)
= 119
2
We calculate the number of diagonals in 17gon.
Ans. The polygon has 119 diagonals.
PROBLEM
How many diagonals has a polygon whose sum of angle measures is 2160◦?
A polygon that has all sides of equal length and all angles of equal measure is called a regular polygon. For example, regular polygons are an
equilateral triangle and a square.
In each regular polygon, all bisectors of angles and perpendicular bisectors
of sides intersect at the same point. Therefore, this point is also the center
of the circle circumscribed about the polygon and circle inscribed in this
polygon.
128
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 128
When calculating the radius length of these circles for an equilateral triangle, a square and a regular hexagon, appropriate formulas can be used
which allow to calculate the equilateral triangle’s height and the length of
square’s diagonal.
√
√
2
6
√
r = 1×a 3 = a 3
r= a
√
R = 2r = a 3
3
√
√
R = 1a 2 = a 2
3
r= a 3
2
2
2
2
R=a
We already know that the sum of the angles in n-gon is (n − 2) × 180◦.
Since all angles of the regular n-gon are equal, we can use the following
theorem.
Theorem
◦
Each angle of a regular n-gon has the measure (n − 2)n× 180 .
EXERCISE F
have?
a) What measure does each of the angles of a regular hexagon
b) How many diagonals does the regular hexagon have?
PROBLEMS
1. a)
Calculate the sum of measures of heptagon’s angles and angles of a 100-gon.
b) How many sides has a polygon in which the sum of the angle measures is 1080◦ ?
c) Is there a polygon in which the sum of all angles is 1000◦ ?
d) In a 10-gon one of the angles is right and the other angles have the same
measure. Calculate this measure.
2. Justify that there is no heptagon in which every two adjacent sides are perpendicular.
PROPERTIES OF POLYGONS. REGULAR POLYGONS
MLR2-1 str. 129
129
3. a)
Calculate the length of the circle circumscribed about an equilateral triangle
with a side of 10.
b) Determine the length of the circle inscribed in an equilateral triangle with a side
of 5.
4. Let’s consider regular polygons inscribed in a circle with center S.
a) How many diagonals of a regular octagon go through point S?
b) Can any diagonal of a regular heptagon pass through point S?
c) What can you say about the number of vertices of a regular polygon in which
at least one diagonal goes through the center of the circle? How many diagonals of
such a regular n-gon go through the center of the circle?
Each of the angles of a regular polygon has 150◦. How many sides does this
polygon have?
5. a)
b) Is there a regular polygon whose every angle is 130◦?
6. For
which n the angle’s measure of a regular n-gon (specified in degrees) is a
number divisible by 10?
7. Chord AB is a side of a regular polygon inscribed in a circle with center S. What
measure each angle of this polygon has if angle’s ASB measure is α?
130
FI GUR E S O N TH E PLANE PAR T 2
MLR2-1 str. 130