Summer School : Algebraic Manipulation and Inequalities
1
1
Introduction
In this part of the summer school we will look at how we handle algebraic expressions. In
particular, we will seek to ‘simplify’ algebraic expressions so that they are more compact
and easier to evaluate. We will then move on to look at polynomials. We will see how to
factorise polynomials and solve equations involving polynomials. We will finish the first
section off by examining the quadratic in detail.
The second section here will deal with inequalities. We will see how inequalities are
solved and how they may be represented as a range or on the number line. We will finish
off by looking at inequalities that require algebraic manipulation to solve.
2
Algebraic Manipulation
2.1
Index Laws
Before we start on algebraic manipulation we have to understand the laws relating to
powers and roots. These laws are referred to as the
Laws for Indices
For numbers a, b > 0 and any numbers x and y we have
1
1. a0 = 1
(Also note a1 = a)
2.
= a−x
x
a
√
x
√
1
4. ax ay = ax+y
3. a y = y ax
(Again note x a = a x )
5.
ax
= ax−y
ay
7.
(ab)x = ax bx
6. (ax )y = axy
These laws have been covered in the introduction lectures of this course and are only
given here as a reminder. Make sure you understand these laws and how to use them !
2.2
Simplifying Expressions
You will have seen in mathematics that expressions can be written in various forms, some
‘simpler’ than others. For example
x2 + 8x + 2x2 + 2x + 18 + 2x − 6 = 3(x + 2)2
What is essential though is that if we evaluate the ‘different’ expressions for any value of
the variable we will get the same answer for each of the different forms.
Summer School : Algebraic Manipulation and Inequalities
2
That is, letting x = 2 in the above expression gives:x2 + 8x + 2x2 + 2x + 18 + 2x − 6 = 22 + 8(2) + 2(22 ) + 2(2) + 18 + 2(2) − 6 = 48
3(x + 2)2 = 3(2 + 2)2 = 3(16) = 48
What we have done is manipulate the given expression, using the Laws for Indices,
into an equivalent one.
Indeed algebraic manipulation consists of writing down a sequence of equivalent expressions, usually with a view to ‘simplifying’ in some way. Each step must make mathematical
sense.
You will have had experience at manipulation throughout your mathematical studies
but may not have realized. We will work through some more complex expressions to see
how manipulation ‘simplifies’.
Example 2.1
3
2
a)
+
x+1 x−2
=
3
x−2
2
x+1
×
+
×
x+1 x−2 x−2 x+1
=
2(x + 1)
3(x − 2)
+
(x + 1)(x − 2) (x + 1)(x − 2)
=
3(x − 2) + 2(x + 1)
(x + 1)(x − 2)
=
3x − 6 + 2x + 2
(x + 1)(x − 2)
=
5x − 4
(x + 1)(x − 2)
x
3x − 2
−
(x − 1)(x + 2) (x − 1)(x + 3)
b)
=
x
x+3
3x − 2
x+2
×
−
×
(x − 1)(x + 2) x + 3 (x − 1)(x + 3) x + 2
=
x(x + 3)
(3x − 2)(x + 2)
−
(x − 1)(x + 2)(x + 3) (x − 1)(x + 2)(x + 3)
=
x2 + 3x − 3x2 − 6x + 2x + 4
4 − x − 2x2
=
(x − 1)(x + 2)(x + 3)
(x − 1)(x + 2)(x + 3)
3
Summer School : Algebraic Manipulation and Inequalities
1
(x2 + x − 3) 3
−
2
(x + 3)2
3(x + 3)(x2 + x − 3) 3
2x + 1
c)
=
=
=
1
2
x+3
(x2 + x − 3) 3
3(x2 + x − 3) 3
×
−
×
2
2
x+3
(x + 3)2
3(x + 3)(x2 + x − 3) 3
3(x2 + x − 3) 3
2x + 1
2
1
(2x + 1)(x + 3)
2
3(x + 3)2 (x2 + x − 3) 3
3(x2 + x − 3) 3 (x2 + x − 3) 3
−
2
3(x + 3)2 (x2 + x − 3) 3
(2x2 + 6x + x + 3) − 3(x2 + x − 3)
2
3(x + 3)2 (x2 + x − 3) 3
−x2 + 4x + 12
=
2
3(x + 3)2 (x2 + x − 3) 3
Reminder:- To check our answers we can put any value of x in and both expressions
should give us the same answer. Try this yourself.
2.3
2.3.1
Polynomials and Rational Functions
Definition of a Polynomial
Pn (x) = a0 xn + a1 xn−1 + a2 xn−2 + ... + an−1 x + an
is called a polynomial of degree n , in the variable x.
The degree of the polynomial is n, the highest power of x present. The numbers
a0 , a1 , a2 , · · · , an−1 , an are real constants, and are called the coefficients. Any of these
coefficients may be zero. If ao = 0 the degree is 1 less. Note polynomials are often written
without the degree as a subscript, that is P (x).
Some examples are:
When
When
When
When
n = 0,
n = 1,
n = 2,
n = 3,
P0 (x) = 6,
P1 (x) = 3x − 8,
P2 (x) = 2x2 + 5x − 3,
P3 (x) = −5x3 + 4x − 1,
which
which
which
which
is
is
is
is
zero degree, or constant.
first degree, or linear.
second degree, or quadratic.
third degree, or cubic.
The variable the polynomial is in does not have to be x. For example,
5w 3 − 7w 2 − 2w + 10
is a polynomial of degree 3 in the variable w.
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Summer School : Algebraic Manipulation and Inequalities
2.3.2
The Value of a Polynomial
The value of Pn (x) when x = α is denoted by Pn (α), and so
Pn (α) = a0 αn + a1 αn−1 + a2 αn−2 + ... + an−1 α + an
Example 2.2
If
P3 (x)
= x3 − 2x2 − x + 1,
P3 (2)
P3 (−5)
P3 (α)
P3 (−x)
P3 (x + 1)
= (2)3 − 2.(2)2 − 2 + 1 = 8 − 8 − 2 + 1 = −1
= (−5)3 − 2.(−5)2 − (−5) + 1 = −125 − 50 + 5 + 1 = −169
= α3 − 2α2 − α + 1
= (−x)3 − 2(−x)2 − (−x) + 1 = −x3 − 2x2 + x + 1
= (x + 1)3 − 2(x + 1)2 − (x + 1) + 1 = x3 + x2 − 2x − 1
So to evaluate a polynomial we replace the variable with the value the variable takes.
2.3.3
Algebraic Operations on Polynomials
We can add, subtract and multiply two polynomials Pn (x), Qm (x), to obtain another polynomial. Addition and subtraction are straightforward, since we simply add or subtract like
powers of x.
Addition and Subtraction of Polynomials
Example 2.3
If P3 (x) = (x3 − 4x2 + 3x − 5), Q3 (x) = (2x3 + 4x2 − 5x − 8) and R2 (x) = (3x2 + 7x + 8)
then,
P3 (x) + Q3 (x)
P3 (x) − R2 (x)
Q3 (x) + R2 (x)
Q3 (x) − 2P3 (x)
= (x3 − 4x2 + 3x − 5) + (2x3 + 4x2 − 5x − 8)
= (x3 − 4x2 + 3x − 5) − (3x2 + 7x + 8)
= (2x3 + 4x2 − 5x − 8) + (3x2 + 7x + 8)
= (2x3 + 4x2 − 5x − 8) − (2x3 − 8x2 + 6x − 10)
Multiplication of Polynomials
= 3x3 − 2x − 13
= x3 − 7x2 − 4x − 13
= 2x3 + 7x2 + 2x
= 12x2 − 11x + 2
Multiplication of Pn (x) by Qm (x) gives a polynomial of degree m + n. The highest power
of x in both Pn (x) and Qm (x) are multiplied together to give the highest power of x in the
new polynomial. Recall the index law xm × xn = xm+n . Multiplication of polynomials can
be done by multiplying out brackets.
Example 2.4
If P3 (x) = 3x3 − x + 1 and Q2 (x) = 2x2 + x − 1, find
a)
2 × P3 (x)
b) 4x × Q2 (x)
c)
P3 (x) × Q2 (x)
Summer School : Algebraic Manipulation and Inequalities
a)
2(3x3 − x + 1)
=
6x3 − 2x + 2
b)
4x(2x2 + x − 1)
=
8x3 + 4x2 − 2x − 4x
c)
(3x3 − x + 1)(2x2 + x − 1) =
=
5
6x5 + 3x4 − 3x3 − 2x3 − x2 + x + 2x2 + x − 1
6x5 + 3x4 − 5x3 + x2 + 2x − 1
Division of Polynomials
Pn (x)
where n ≥ m, and where Pn , Qm have no common factor,
If we want to evaluate
Qm (x)
the result is not in general another polynomial, but another polynomial of degree n − m,
since xn /xm = xn−m , plus a remainder of degree < m. Note here the remainder may be
zero.
Example 2.5
Divide 3x4 + 4x3 − 5x2 + 2x + 1 by x2 − 2x − 3. This is similar to long division in
arithmetic.
3x2
10x
24
x2 −2x −3 ) 3x4
4x3
−5x2
2x
1
4
3
2
− 3x −6x
−9x
10x3
4x2
2x
1
3
2
−
10x −20x −30x
24x2
32x
1
−
24x2 −48x −72
80x
73
So we have (3x4 + 4x3 − 5x2 + 2x + 1) = (x2 − 2x − 3)(3x2 + 10x + 24) + 80x + 73 (?)
We can check our division is correct by evaluating both sides of (?) for ANY value of x.
Take x = 2,
3x4 + 4x3 − 5x2 + 2x + 1
= 3(2)4 + 4(2)3 − 5(2)2 + 2(2) + 1
= 48 + 32 − 20 + 4 + 1 = 65
and
(x2 − 2x − 3)(3x2 + 10x + 24) + 80x + 73
= ((2)2 − 2(2) − 3)(3(2)2 + 10(2) + 24) + 80(2) + 73
= (4 − 4 − 3)(12 + 20 + 24) + 160 + 73 = (−3)(56) + 233 = −168 + 233 = 65
Example 2.5 can be written
80x + 73
3x4 + 4x3 − 5x2 + 2x + 1
= 3x2 + 10x + 24 + 2
.
2
x − 2x − 3
x − 2x − 3
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Summer School : Algebraic Manipulation and Inequalities
Written out in full the problem is: dividing 3x4 + 4x3 − 5x2 + 2x + 1 (the dividend) by
x2 − 2x − 3 (the divisor), we obtain 3x2 + 10x + 24 (the quotient), and the remainder
80x + 73.
So we started with an ‘improper’ rational expression (numerator higher degree than
denominator) and have written it as a polynomial plus a ‘proper’ rational expression. The
result we get is unique if we ensure that the degree of the remainder is less than the degree
of the divisor.
2.3.4
The Remainder Theorem
This important theorem states that when P (x) is divided by the linear polynomial x − α,
the remainder R will be a constant equal to P (α).
Proof :We have
P (x) = (x − α)Q(x) + R.
(??)
R must be of lower degree than x − α, and so it must be a constant.
Putting x = α into (??) gives
P (α) = 0 × Q(α) + R
i.e. R = P (α).
This result does not give us Q(x) which can be obtained by division.
Example 2.6
The remainder when P (x) = x3 − 4x2 + 5x − 3 is divided by x − 2 is
P (2) = 23 − 4 × 22 + 5 × 2 − 3 = −1.
Since we are dividing by a linear factor, (x − 2) we can use ‘synthetic division’ to find
the other factor.
2
+
1 -4 5
↓ 2 -4
1 -2 1
-3
2
-1 =R
The last row gives us the coefficients of the quotient, which we know is of degree one
less than P (x), and the remainder R. So here the quotient is x2 − 2x + 1, with remainder
−1 and
1
x3 − 4x2 + 5x − 3
= x2 − 2x + 1 −
x−2
x−2
i.e. x3 − 4x2 + 5x − 3 = (x − 2)(x2 − 2x + 1) − 1.
Summer School : Algebraic Manipulation and Inequalities
2.3.5
7
Factorising Polynomials
Here we want to write a given polynomial as a product of its factors. A factor of a
polynomial is another polynomial of smaller degree that divides into the original polynomial
with no remainder.We usually look for linear factors (a linear factor involves x to the power
1).
We can get
Pn (x) = a0 (x − α1 )(x − α2 ) · · · (x − αn )
This result supposes that all the factors of Pn are linear with real coefficients. Note the
subtraction in the bracket.
We can also have
Pn (x) = a0 (x − α1 )(x − α2 ) · · · (x − αr )Qn−r (x)
To factorise we need to get trial factors and check with the Remainder Theorem.
Suppose that
Pn (x) = a0 xn + a1 xn−1 · · · an−1 x + an
where, ar , (r = 0, 1 · · · n) are real numbers.
A hint to help us to select trial factors is to try factors of the form x − α where α is a
factor of an .
Example 2.7
Find the factors of P (x) = x4 − 4x3 − 4x2 − 4x − 5.
Note that the possible integer values of α are the factors of 5, i.e. ±1, ±5.
Try x − 1: P (1) = 1 − 4 − 4 − 4 − 5 6= 0
Try x + 1: P (−1) = 1 + 4 − 4 + 4 − 5 = 0
So (x + 1) is a linear factor and P = (x + 1)Q3 (x). To find Q3 (x) we use our synthetic
division.
-1
+
1
↓
1
-4 -4 -4
-1 5 -1
-5 1 -5
-5
5
0 =R
Now try for factors of Q(x) = x3 − 5x2 + x − 5. Note that we do not try x − 1 again
Try x + 1: Q(−1) = −1 − 5 − 1 − 5 6= 0
Try x − 5: Q(5) = 53 − 5.52 + 5 − 5 = 0
Summer School : Algebraic Manipulation and Inequalities
8
So (x − 5) is a factor, and
Q(x) = (x − 5)(x2 + 1). The quadratic here is found using synthetic division.
The remaining factor x2 + 1 is irreducible (that is, cannot be factorised any further),
and so
P (x) = (x + 1)(x − 5)(x2 + 1).
2.3.6
Solving Polynomial Equations
When we are asked to solve a polynomial equation we follow the following steps.
1. Gather everything to the LHS so we have Pn (x) = 0
2. Factorise the polynomial fully
3. Solve for each factor in turn
This process is best seen in an example.
Example 2.8
Solve for x, x3 − 6x2 + 6x − 3 = 3 − 5x
⇒ x3 − 6x2 + 11x − 6 = 0
To factorise we note possible values of α in (x − α) are ±1, ±2, ±3 and ± 6.
Try x + 1: P (−1) = −1 − 6 − 11 − 6 6= 0
Try x − 1: P (1) = 1 − 6 + 11 − 6 = 0
So (x − 1) is a factor. To find the other factors we use synthetic division.
1
+
1 -6 11
↓ 1 -5
1 -5 6
-6
6
0 =R
Now factorise x2 − 5x + 6. Note that we do not try x − 1 again
Try x + 1: Q(−1) = 1 + 5 + 6 6= 0
Try x − 2: Q(2) = 4 − 10 + 6 = 0
So (x − 2) is a factor, and (x − 3) is the last factor which is found using synthetic division
or by inspection.
So we have
x3 − 6x2 + 11x − 6 = 0
⇒ (x − 1)(x − 2)(x − 3) = 0
⇒ (x − 1) = 0
⇒ x=1
(x − 2) = 0 (x − 3) = 0
x=2
x=3
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Summer School : Algebraic Manipulation and Inequalities
2.3.7
Quadratics
A quadratic is a polynomial of degree 2 i.e. ax2 + bx + c, where a 6= 0. Often we desire to
‘complete the square’ in a quadratic. That is, given
ax2 + bx + c
we rewrite in the form
k(x + l)2 + m
This has many uses which we will see in a minute but first we will see how to manipulate
ax2 + bx + c into k(x + l)2 + m.
Example 2.9
a)
x2 + 4x + 6 =
(x + 2)2 − (2)2 + 6
=
−x2 + 8x − 5
=
−(x − 4)2 + 16 − 5 =
=
3x2 + 6x − 9
=
2
3(x + 1) − 3 − 9 =
=
−2x2 + 8x − 8
=
2
−2(x − 2) + 8 − 8 =
b)
c)
d)
= (x + 2)2 − 4 + 6 =
−(x2 − 8x) − 5 =
11 − (x − 4)2
3(x2 + 2x) − 9
3(x + 1)2 − 12
(x + 2)2 + 2
−[(x − 4)2 − (−4)2 ] − 5
= 3[(x + 1)2 − (1)2 ] − 9
−2(x2 − 4x) − 8 = −2[(x − 2)2 − (−2)2 ] − 8
−2(x − 2)2
There are two main uses of this technique, 1) Min/Max problems and 2) Finding roots.
Min/Max Problems
Take example a) from above x2 + 4x + 6 = (x + 2)2 + 2. We can see that the smallest
value this can take is 2, since (x + 2)2 ≥ 0. What’s more, we know this minimum occurs
when x = −2. We can get this information from a graph as well :y=x2 +4x+6
12
10
8
6
4
2
−5
−4
−3
−2
−1
0
1
2
Summer School : Algebraic Manipulation and Inequalities
10
Similarly with example b) −x2 + 8x − 5 = 11 − (x − 4)2 , we can conclude that the
maximum this function takes is 11 when x = 4. Try drawing the graph to confirm this.
Finding Roots
If we are asked to find the roots of ax2 + bx + c = 0 we can use completing the square
to do this.
Take
⇒
⇒
⇒
3x2 + 6x − 9 = 0
3(x + 1)2 = 12
x + 1 = ±2
x = −3 OR x = 1
⇒ 3(x + 1)2 − 12 = 0 From c)
⇒ (x + 1)2 = 4
⇒ x = −1 ± 2
You may have seen the quadratic formula before,
x=
−b ±
√
b2 − 4ac
2a
and this if found by completing the square in ax2 + bx + c = 0 as follows
b
c
ax2 + bx + c = 0
⇒ x2 + x + = 0
a
a
⇒
b 2 c
b2
+ − 2 =0
x+
2a
a 4a
⇒
x+
⇒
√
b
± b2 − 4ac
x+
=
2a
2a
b 2 b2 − 4ac
=
2a
4a2
b2
b 2
c
= 2−
⇒
x+
2a
4a
a
√
b
± b2 − 4ac
√
⇒ x+
=
2a
4a2
√
−b ± b2 − 4ac
⇒ x=
2a
So we can use this formula to find the roots of any quadratic.
Example 2.10
Find the roots of
4x2 − 8x + 8 = 6.
The first thing we have to do is put this in the form
ax2 + bx + c = 0 ⇒ 4x2 − 8x + 2 = 0
From this we can see we have a = 4, b = −8 and c = 2 which we substitute into the
formula to get:-
Summer School : Algebraic Manipulation and Inequalities
⇒
⇒
⇒
x=
−b ±
x=
8±
x= 1±
x= 1±
√
√
√
√
b2 − 4ac
2a
64 − 32
8
16 × 2
8
2
2
⇒ x=
8±
p
(−8)2 − 4(4)(2)
2(4)
√
32
8
√
4 2
⇒ x = 1±
8
⇒ x = 1±
11
12
Summer School : Algebraic Manipulation and Inequalities
3
Inequalities
Given that a and b are any two real numbers, inequalities are defined as follows
a<b
a>b
a≤b
a≥b
···
···
···
···
‘a
‘a
‘a
‘a
is
is
is
is
less than b’
greater than b’
less than or equal to b’
greater than or equal to b’
(Reminder : Points to the Left ⇒ Lessthan).
3.1
Solving Inequalities
The way to solve an inequality is very similar to the way we solve an equation, BUT
THERE IS ONE IMPORTANT DIFFERENCE !
We can add the same number to both sides
(add 2 to both sides)
x−2<4
x<6
We can subtract the same number to both sides
(subtract 6 from both sides)
x + 6 < 10
x<4
We can multiply or divide both sides by a POSITIVE NUMBER
(divide both sides by 3)
3x < 18
x<6
BUT when we multiply or divide both sides by a NEGATIVE NUMBER, the inequality
is REVERSED.
(divide both sides by -3)
−3x < 18
x > −6
(Note that the inequality has reversed)
3.2
Some Simple Problems and the Number Line
The interval (a, b) means ‘all values in between a and b but NOT including a or b’. On the
number line this is shown as :-
a
b
13
Summer School : Algebraic Manipulation and Inequalities
The interval [a, b] means ‘all values in between a and b INCLUDING a and b’. On the
number line this is shown as :-
a
b
So, for example, (3, 8) is represented as :-
3
8
and [4, 10) would be represented as :-
4
10
Note that this interval INCLUDES 4 but EXCLUDES 10.
We now move on to solving some simple inequalities.
Example 3.1
(multiply both sides by 2)
(subtract 1 from both sides)
1
(x + 1) ≤ 6
2
x + 1 ≤ 12
x ≤ 11
This can be represented on the number line as :-
11
which can also be written as (−∞, 11].
NOTE :- The sign for infinity ‘∞’ is always closed with a round bracket since ∞ is not
a real number and therefore cannot be included in the interval.
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Summer School : Algebraic Manipulation and Inequalities
Example 3.2
(divide both sides by -3)
(subtract 4 from both sides)
(multiply both sides by -1)
−3(4 − x) < 12
4 − x > −4
−x > −8
x<8
This can be represented on the number line as :-
8
which can also be written as (−∞, 8).
Note that ‘8’ is closed by a round bracket i.e. ‘8’ is NOT included in the interval.
Example 3.3 (A little bit harder)
1 2
(x − 4x + 3) < 0
52
(multiply both sides by 5)
x − 4x + 3 < 0
(factorise the quadratic)
(x − 1)(x − 3) < 0
WARNING :- This does NOT mean x − 1 < 0 and x − 3 < 0 as we would do with
equality !
We see that the product (x − 1)(x − 3) is zero at x = 1 and x = 3 but what happens
elsewhere ? Well, we draw up the following table to find out :(x − 1)
(x − 3)
(x − 1)(x − 3)
→ 1 →
- 0 +
- - + 0 -
3 →
+ +
0 +
0 +
The first row shows what sign the factor (x − 1) takes for x values ranging from less
than 1 to greater than 3.
The second row shows what sign the factor (x − 3) takes.
The third row shows what sign (x − 1)(x − 3) takes by multiplying the signs for each
factor. That is :(−)(−)
(0)(−)
(+)(−)
(+)(0)
(+)(+)
= +
= 0
= −
= 0
= +
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Summer School : Algebraic Manipulation and Inequalities
This is the row we are interested in ! We require (x − 1)(x − 3) < 0. From the third row
we see the negative sign lies in the range x > 1 , x < 3 or (1, 3) which can be represented
on the number line as :-
1
3.3
3
Further Examples
Now we have the basics, we are ready to move on and try some fairly complex inequalities.
Example 3.4 (A cubic)
(x + 3)(2x + 1)(3x − 2) > 0
We see that (x + 3)(2x + 1)(3x − 2) is zero when either
x + 3 = 0 2x + 1 = 0 3x − 2 = 0
1
2
x = −3
x=−
x=
2
3
We therefore draw up our table as follows:1
→ -3 → −
2
(x + 3)
- 0 + +
(2x + 1)
0
(3x − 2)
(x + 3)(2x + 1)(3x − 2) - 0 +
0
→
+
+
-
2
3
+
+
0
0
→
+
+
+
+
We require (x + 3)(2x + 1)(3x − 2) > 0. So we look for the positive signs in the bottom
−1
2
row. We see that we require x > −3 , x <
and x >
which can be written as
2
3
2
1
x ∈ (−3, − ) and x ∈ ( , ∞) and can be represented on the number line as :2
3
-3
-1/2
NOTE :- x ∈ (a, b) means ‘x’ is in the set (a, b).
2/3
Summer School : Algebraic Manipulation and Inequalities
3.4
16
Examples Involving Fractions
These are handled in a similar way to the previous examples but may require slightly more
manipulation.
Example 3.5
(x2
(factorise the denominator)
(x − 2)
>0
− 4x + 3)
(x − 2)
>0
(x − 3)(x − 1)
When we get to this stage we examine the factors (x − 2),(x − 3) and (x − 1) (as before)
which we see are equal to zero when
x−2=0 x−3=0
x=2
x=3
The table now
→
(x − 1)
(x − 2)
(x − 3)
fraction -
looks like
1 →
0 +
nd +
x−1=0
x=1
:2
+
0
0
→ 3
+ +
+ +
0
- nd
→
+
+
+
+
Note :- ‘nd’ stand for ‘not defined’ and occurs when we divide by 0.
(x − 2)
> 0, which means the solution is x ∈ (1, 2) and x ∈ (3, ∞)
(x − 1)(x − 3)
which can be represented as :We require
1
2
3
The next example is similar to an exam type question.
Example 3.6
2
1
≥
(x − 2)
(3x + 1)
The basic idea is to bring everything to the left hand side, combine the fractions to
become one and factorise the numerator and denominator. Then we simply examine the
factors as in the previous examples.
Summer School : Algebraic Manipulation and Inequalities
(bring everything to LHS)
2
1
−
≥0
(x − 2) (3x + 1)
(combine fractions)
1(3x + 1) − 2(x − 2)
≥0
(x − 2)(3x + 1)
(simplify)
3x − 2x + 1 + 4
≥0
(x − 2)(3x + 1)
17
(x + 5)
≥0
(x − 2)(3x + 1)
The factors (x + 5),(x − 2) and (3x + 1) are equal to zero when
x + 5 = 0 x − 2 = 0 3x + 1 = 0
1
x = −5
x=2
x=−
3
We now have a table like :(x + 5)
(3x + 1)
(x − 2)
fraction
1
3
+
0
nd
→ -5 → −
-
0
0
+
+
→
+
+
-
2
+
+
0
nd
→
+
+
+
+
1
(x + 5)
≥ 0, hence the solutions are x ∈ [−5, − ) and x ∈ (2, ∞)
(x − 2)(3x + 1)
3
which can be represented as :We require
-5
-1/3
2
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Summer School : Algebraic Manipulation and Inequalities
3.5
The Number Line Revisited
Suppose we have the interval x ≥ 2, x ≤ 5
2
5
This can also be given by [2, 5].
Using inequalities, this interval can be written in a neat form. We have :(bring the 2 to the LHS and the x to the RHS)
(multiply by -1)
x≥2
−2 ≥ −x
2≤x
So x ≥ 2 is equivalent to 2 ≤ x (obvious ?).
Then 2 ≤ x , x ≤ 5 can be written as 2 ≤ x ≤ 5. In general we have:-
a < x < b · · · x ∈ (a, b) · · · x is in the interval (a, b) not including a and b
a
b
a ≤ x ≤ b · · · x ∈ [a, b] · · · x is in the interval [a, b] including a and b
a
b
Example 3.7
Sketch the interval −3 < x ≤ 2 on the number line.
-3
which can also be written as (−3, 2].
2
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Summer School : Algebraic Manipulation and Inequalities
3.6
Modulus
Definition :Define |x|, the ‘modulus of x’ as
x, x ≥ 0
|x| =
−x, x ≤ 0
In plain English, |x| is the absolute (positive) value of x. You can think of |x| as being
the length of x.
e.g. |10| = 10 and | − 10| = 10.
3.7
Inequalities With Modulus
Consider the inequality |x| ≤ 4. This can be represented on the number line as :-
-4
0
4
But this can also be written as x ≥ −4, x ≤ 4, which (from before) can be written as
−4 ≤ x ≤ 4. So there are lots of different ways the exact same range can be expressed.
What about |x − 3| < 7 ?
Using the same trick as above, this means
(adding 3 to both)
which we can write as
x − 3 > −7
x > −4
x−3<7
x < 10
−4 < x < 10
So if we have an inequality of the form
|x − c| < r
this can be written as
−r < x − c < r
⇒ −r + c < x < r + c
Suppose we would like to go the other way
a < x < b ⇒ |x − c| < r ?
Summer School : Algebraic Manipulation and Inequalities
Example 3.8 (The Last One !)
3<x<7
From above,
−r + c < x < r + c
⇒
−r + c = 3
⇒
r+c=7
Add together to get
2c = 10
⇒
c=5
Sub back in c = 5
−r + 5 = 3
r=2
So we have 3 < x < 7 is equivalent to |x − 5| < 2.
Next :- Geometry with Dr. Coles
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