 Monika V Sikand
Light and Life Laboratory
Department of Physics and Engineering physics
Stevens Institute of Technology
Hoboken, New Jersey, 07030.
OUTLINE
Square Roots
Complex Numbers
The Quadratic Formula and the Discriminant
A quadratic function has a form
y = ax2 + bx + c
where a ≠ 0.
The graph of a quadratic function is Ushaped and is called parabola.
y = x2
vertex
The lowest or highest point on the graph of
a quadratic function is called the vertex.
The graphs of y = x2 and y = -x2 are
symmetric about the y-axis, called the axis
of symmetry.
y = -x2
Axis of
symmetry
FUNCTION
 The parabola opens up if a&gt;0 and
opens down if a&lt;0.
 The parabola is wider than the
graph of y = x2 if |a| &lt; 1 and
narrower than the graph of y = x2 if
|a| &gt; 1.
The x-coordinate of the vertex is
-b/(2a).
The axis of symmetry is the vertical
line x = -b/(2a).
y = x2
vertex
y = -x2
Axis of
symmetry
EXAMPLE
Graph y = 2x2 -8x +6
Solution: The coefficients for this function are
a = 2, b = -8, c = 6.
Since a&gt;0, the parabola opens up.
The x-coordinate is: x = -b/2a,
x = -(-8)/(2(2))
x=2
The y-coordinate is: y = 2(2)2-8(2)+6
y = -2
Hence, the vertex is (2,-2).
EXAMPLE(contd.)
Axis of symmetry
Draw the vertex (2,-2) on graph.
Draw the axis of symmetry x=-b/(2a).
(4,6)
(0,6)
Draw two points on one side of the
axis of symmetry such as (1,0) and
(0,6). How were these points chosen?
Use symmetry to plot two more points
such as (3,0), (4,6).
Draw parabola through the plotted
points.
(1,0)
(3,0)
(2,-2)
y
x
EQUATION
y = a(x - h)2 + k
The vertex is (h,k).
The axis of symmetry is x = h.
FUNCTION IN VERTEX FORM
Axis of symmetry
Example y = -1/2(x + 3)2 + 4
where a = -1/2, h = -3, k = 4. Since a&lt;0,
the parabola opens down.
(-3,4)
To graph a function, first plot the vertex
(h,k) = (-3,4).
(-5,2)
Draw the axis of symmetry x = -3
(-1,2)
Plot two points on one side of it, such as
(-1,2) and (1,-4).
x
Use the symmetry to complete the
graph.
(-7,-4)
y
(1,-4)
INTERCEPT FORM OF
y = a(x - p)(x - q)
The x intercepts are p and q.
The axis of symmetry is halfway
between (p,0) and (q,0).
FUNCTION IN INTERCEPT FORM
Axis of symmetry
Example y = -(x + 2)(x - 4).
where a = -1, p = -2, q = 4. Since a&lt;0
the parabola opens down.
(1,9)
To graph a function, the x-intercepts
occur at (-2,0) and (4,0).
Draw the axis of symmetry that lies
halfway between these points at x = 1.
So, the x - coordinate of the vertex is
x = 1 and the y - coordinate of the
vertex is: y = -(1 + 2)(1 - 4)= 9.
(-2,0)
(4,0)
x
y
EQUATION IN STANDARD FORM
(1). y = -(x + 4)(x - 9)
= -(x2 - 9x + 4x - 36)
= -(x2 - 5x -36)
= -x2 + 5x + 36
(2). y = 3(x -1)2 + 8
= 3(x -1)(x - 1) + 8
= 3(x2 - x - x + 1) + 8
= 3(x2 - 2x + 1) + 8
= 3x2 - 6x + 3 + 8
= 3x2 - 6x + 11
REAL LIFE
Researchers conducted an experiment to determine
temperatures at which people feel comfortable. The
percent of test subjects who felt comfortable at
temperature x( in degrees Fahrenheit) can be modeled
by:
y
=
-3.678x2
+
527.3x
–
18,807
a. What temperature made the greatest percent of test
subjects comfortable?
b. At that temperature , what percent felt comfortable?
SOLUTION
Since a = -3.678 is negative, the graph of the quadratic function open
down and the function has a maximum value. The maximum value
occurs at:
b
527.3
x 
 72
2a
2(3.678)
The corresponding value of y is:

a.
b.

y  3.678(72)2  527.3(72) 18, 807  92
Hence, The temperature that made the greatest percent of test
At that temperature about 92% of the subjects felt comfortable.
REAL LIFE EXAMPLE
The Golden Gate Bridge in San Francisco has two towers that
rise 500 feet above the road and are connected by suspension
cables as shown. Each cable forms a parabola with equation
1
y
(x  2100)2  8
8940
where x and y are measured in feet.
a. What is the distance d between the towers?
b. What is the height l above the road of a cable at its lowest
point?
GOLDEN GATE BRIDGE
y
500ft
200ft
Not drawn to scale
l
d
x
SOLUTION
1
(x  2100)2  8
8960
1
y
(x 2  4200x  4410000)  8
8960
1
4200
4410000
2
y
x 
x(
 8)
8960
8960
8960
b 4200
1
x
(
)(
)
1
2a 8960 2 
8960
x  2100

y
1
(x  2100)2  8
8960
1
y
(2100  2100)2  8
8960
0
y
8
8960
y8
y
Hence the vertex of the parabola is (2100,8)
SOLUTION(contd.)
a. The vertex of the parabola is (2100,8), so the
cable’s lowest point is 2100 feet from the left tower
shown above. Since the heights of the two tower’s
are the same, the symmetry of the parabola implies
that the vertex is also 2100 feet from the right tower.
Therefore the towers are d = 2(2100) = 4200 feet
apart.
b. The height l above the road of a cable at its lowest
point is the y-coordinate of the vertex. Since the
vertex is (2100,8), this height is l= 8 feet.
EQUATION BY FACTORING
EXPRESSION
The expression x2 + bx + c is a trinomial because it
has three terms. We can use factoring to write it as
product of two terms or binomials such as
x2 + bx + c = (x+m)(x+n) = x2 + (m+n)x + mn
Example:
x2 + 8x + 15 = (x + 3)(x + 5)
EXAMPLE
Problem: Factor x2 - 12x - 28
Solution:
x2 - 12x - 28 = (x+m)(x+n) where mn = -28 and m+n = -12
Factors
of -28
Sum of
factors
-1,28 1,-28 -2,14 2,-14
-4,7
4,-7
27
3
-3
-27
12
-12
The table shows that m = 2 and n = -14.
So, x2 - 12x - 28 = (x + 2)(x - 14)
SPECIAL FACTORING
PATTERNS
1. Difference of two squares: a2 - b2 = (a+b)(a-b)
Example: x2 - 9 = (x+3)(x-3)
2. Perfect square Trinomial: a2 + 2ab + b2 = (a+b)2
Example: x2 + 12x +36 = (x+6)2
3. Perfect square Trinomial: a2 - 2ab + b2 = (a-b)2
Example: x2 -8x +16 = (x-4)2
FACTORING MONOMIALS FIRST
Monomial is an expression that has only one term.
a. 5x2 - 20 = 5(x2 -4)
= 5(x+2)(x-2)
b. 6p2 + 15p + 9 = 3(2p2 + 5p + 3)
= 3(2p + 3)(p + 1)
EQUATIONS
Solve:
x 2  3x 18  0
Solve:
2t 2 17t  45  3t  5
(x  6)(x  3)  0
x6 0
2t 2  20t  50  0
x  6
or
x  3  0,
x3
The solutions are -6 and
 3.
t 2 10t  25  0
(t  5)2  0
t 50
t 5
The solution is 5.
REAL LIFE EXAMPLE
You have made a rectangular stained glass window
that is 2 feet by 4 feet. You have 7 square feet of clear
glass to create a border of uniform width around the
window. What should the width of the border be?
x
x
x
x
4
x
x
x 2 x
2+2x
4+2x
SOLUTION
Let the width of the border be = x
Area of the border = 7
Area of the border and window = (2+2x)(4+2x)
Area of the window = 24 = 8
Area of border = Area of border &amp; window - Area of window
7 = (2+2x)(4+2x) - 8
0 = 4x2 + 12x -7
0 = (2+7x)(2x-1)
2+7x = 0
or 2x-1 = 0
x = -3.5
o x = 0.5
Rejecting negative value, -3.5. Hence the border’s width is 0.5ft
ZERO PRODUCT PROPERTY
Let A and B be real numbers or algebraic
expressions. If AB=0, then either A = 0 or B = 0
Solve (a) x2 + 3x -18 = 0
(x + 6)(x - 3) = 0.
Hence either x + 6 = 0 or x - 3 = 0.
The solutions are
x = -6 or x = 3
ZERO PRODUCT PROPERTY
Let A and B be real numbers or algebraic
expressions. If AB=0, then either A = 0 or B = 0
Solve (a) 2t2 - 17t + 45 = 3t - 5
2t2 - 20t + 50 = 0
t2 - 10t + 25 = 0
(t - 5)2 = 0
t-5 =0
t=5
Hence the solutions is 5.
FUNCTIONS
Find zeros of y = x2 -x -6
Solution:
y = x2 - 3x + 2x - (32)
y = x(x-3) + 2(x-3)
y = (x + 2)(x - 3)
Hence the zeros of the function are -2 and 3.
EQUATIONS BY FINDING
SQUARE ROOTS
SQUARE ROOT
9 3
 9  3.

PROPERTIES OF SQUARE ROOTS
(a&gt;0, b&gt;0)
PRODUCT PROPERTY:
ab  a. b
QUOTIENT PROPERTY:
a
a

b
b
SIMPLIFY
(a). 24
 4. 6
2 6
7
(b).
2
7
2


2
2
14

2
Solve:
Solve:
2x 2 1  17
x2  8
1
(x  5)2  7
3
(x  5)2  21
x  8
(x  5)   21
2x 2  16
x  5  21
x  2 2
Hence the solutions are:

2 2
and
2 2
Hence the solutions are:

5  21
and
5  21
REAL LIFE EXAMPLE
A stunt man working on the set of a movie is to fall
out of a window 100 feet above the ground. For the
stunt man’s safety, an air cushion 26 feet wide by
30 feet long by 9 feet high is positioned on the
ground below the window.
a. For how many seconds will the stunt man fall before
he reaches the cushion?
b. A movie camera operating at a speed of 24 frames
per second records the stunt man’s fall. How many
frames of film show the stunt man falling?
SOLUTION
a. The stunt man’s initial height is ho = 100 feet, so his height as a
function of time can be modeled by function h = -16t2 + 100. Since the
height of the cushion is 9 feet above the ground, the the time taken by
the stunt man to reach the cushion is:
h = -16t2 + 100
9 = -16t2 + 100
-91 = -16t2
91/16 = t2 or t ≈ 2.4.
b. Thus, it takes about 2.4 seconds for the stunt man to reach the cushion.
c. The number of frames of film that show the stunt man falling is given by
the product (2.4sec)(24 frames/sec), or about 57 frames.
COMPLEX NUMBERS
COMPLEX NUMBER
A complex number written in standard form is a
number a+bi where a and b are real numbers.
The number a is the real part of the complex
number and number bi is the imaginary part.
If b≠0, then a+bi is an imaginary number. If a=0 and
b≠0, then a+bi is a pure imaginary number.
A complex plane has a horizontal axis called the
real axis and a vertical axis called the imaginary axis.
THE SQUARE ROOT OF A
NEGATIVE NUMBER
1. If r is a positive real number, then
r  i r
where i is the imaginary unit defined as i  1

2. By property (1), it follows that
(i r )  r
2

Complex Number Cycle
EQUATION
Solve:
3x 2  10  26
Solution
3x 2  10  26
3x 2  36
Hence, the solutions are
2i 3 and 2i 3
x  12
2
x   12
x  i 12
x  2i 3



PLOTTING COMPLEX NUMBERS
y
Plot 2-3i in the complex
plane.
To plot 2-3i , start at the
origin, move 2 units to the
right and then move 3 units
down.
0
x
2-3i
COMPLEX NUMBERS
Sum of complex numbers:
Difference of complex numbers:
(a+bi) + (c+di) = (a+c) + i(b+d)
(a+bi) - (c+di) = (a-c) + i(b-d)
Example:
(4-i) + (3+2i) = (4+3) + i(-1+2)
=7+i
Example:
(7-5i) - (1-5i) = (7-1) + i(-5+5)
= 6 + 0i
=6
MULTIPLYING THE COMPLEX
NUMBERS
Write the expression as a complex number in standard form.
a. 5i(-2+i) = -10i + 5i2
= -10i + 5(-1)
= -5-10i
b. (7-4i)(-1+2i) = -7 + 14i + 4i - 8i2
= -7 + 18i - 8(-1)
= 1 + 18i
DIVIDING COMPLEX NUMBERS
5  3i
Write the quotient
1 2i
in standard form.
Solution:
Multiply the numerator and denominator by the complex
conjugate of the denominator.

5  3i 1 2i
5 10i  3i  6i 2


1 2i 1 2i 1 2i  2i  4i 2
113i
1 13
  i
5
5 5
ABSOLUTE VALUES OF COMPLEX
NUMBER
Find the absolute value of each complex number.
(a)3 4i  3  4  25  5
2
2
(b)2i  0  (2i)  0 2  (2)2  2
(c)1 5i  (1)  5  26  5.10
2
2
AND THE DISCRIMINANT
Let a, b, and c be real numbers such that a≠0. The
solutions of the quadratic equation ax2 + bx +c = 0
are:
b  b  4ac
x
2a
2

WITH TWO REAL SOLUTIONS
Solve
2x 2  x  5
Solution :
The solutions are:
2x 2  x  5  0
x
b  b  4ac
x
2a
and
1 12  4(2)(5)
x
2(2)
x
2
x
1 41
4

1 41
 1.35
4
1 41
 1.85
4
WITH ONE REAL SOLUTIONS
Solve
x 2  x  5x  9
Solution :
x 2  x  5x  9
x 2  6x  9  0
6  (6)2  4(1)(9)
x
2(1)
6 0
2
x3
x
Hence, the solution is 3.
WITH TWO IMAGINARY SOLUTIONS
Solve
x 2  2 x  2
Solution
x 2  2 x  2
x 2  2 x  2  0
x
2 
2 2  4(1)(2)
2(1)
2  4
2
2  2i
x
2
x  1 i
x
The solutions are:
1+i and 1-i
DISCRIMINANT
In the quadratic formula, the expression b2-4ac
under the radical sign is called the discriminant
of the associated equation ax2 + bx + c = 0.
b  b  4ac
x
2a
2
NUMBER AND TYPE OF SOLUTIONS
Consider the quadratic equation ax2 + bx + c = 0.
If b2-4ac &gt; 0, then the equation has two real solutions.
If b2-4ac = 0, then the equation has one real solutions.
If b2-4ac &lt; 0, then the equation has two imaginary
solutions.
EXAMPLE: TWO REAL SOLUTIONS
Solve
x 2  6x  8  0
Discri min ant :
b 2  4ac  (6)2  4(1)(8)  4
Solutions :
b  b 2  4ac
x
2a
6  (6)2  4(1)(8)
x
2
6 4 62
 31  4, 2

x
2
2
Hence there are two real
solutions: 4,2
EXAMPLE: ONE REAL SOLUTION
Solve
x 2  6x  9  0
Discri min ant :
b 2  4ac  (6)2  4(1)(9)  0
Solutions :
b  b 2  4ac
x
2a
6  (6)2  4(1)(9)
x
2
6 0 6
 3
x
2
2
Hence, there is one real solution: 3
EXAMPLE: TWO IMAGINARY
SOLUTIONS
Find the discriminant of the quadratic equation and give the number
and type of solutions of the equation.
Solve
x 2  6x 10  0
Discri min ant :
b 2  4ac  (6)2  4(1)(10)  4
Solutions :
b  b 2  4ac
x
2a
6  (6)2  4(1)(10)
x
2
6  4 6  2 1
 3 i

x
2
2
Hence there are two
imaginary solutions:
3+i and 3-i
REAL LIFE EXAMPLE
A baton twirler tosses a baton into the air. The baton
leaves the twirler’s hand 6 feet above the ground and
has an initial vertical velocity of 45 feet per second.
The twirler catches the baton when it falls back to a
height of 5 feet. For how long is the baton in the air?
Solution:
Since the baton is thrown, we use the model
h = -16t2 + vot + ho where vo = 45, ho = 6, h = 5.

SOLUTION(contd.)
h  16t 2  vot  ho
5  16t 2  45t  6
0  16t 2  45t 1
45  2089
t
32
t  0.022
or
t  2.8
Rejecting the negative solution, the
baton is in the air for about 2.8
seconds.
GRAPHING AND SOLVING
INEQUALITY IN TWO VARIABLES
y &lt; ax2 +bx +c
y &gt; ax2 +bx +c
y ≤ ax2 +bx +c
y ≥ ax2 +bx +c
 Draw parabola with equation y = ax2 +bx +c . Make
the parabola dashed for inequalities with &lt; or &gt; and
solid for inequalities with ≤ or ≥ .
 Choose a point (x,y) inside the parabola and check
whether the point is a solution of the inequality.
 If a point (x,y) is a solution, shade the region inside
the parabola. If it is not the solution, shade the region
outside the parabola.
EXAMPLE
y
Graph y &gt; x2 -2x -3
Solution:
Graph y = x2 -2x -3. Since the
inequality is &gt;, make parabola dashed.
Test the point (1,0) inside the
parabola.
y &gt; x2
-2x -3
0 &gt; 12-2(1)-3
0 &gt; -4
So, (1,0) is a solution of the inequality.
Shade the region inside the parabola.
1
0


(1,0) 4 x
REAL LIFE EXAMPLE
You are building a wooden bookcase. You want
to choose a thickness d(in inches) for the
shelves so that each is strong enough to support
60 pounds of books without breaking. A shelf can
safely support a weight of W ( in pounds)
provided that
W ≤ 300d2
a. Graph the given inequality
b. If you make each shelf 0.75 inch thick, can it
support a weight of 60 pounds?
SOLUTION
a.
b.
c.
Graph W = 300d2 for non negative values of
d. Since the inequality symbol is ≤ make the
parabola solid. Test a point inside the
parabola, such as (0.5, 240).
W ≤ 300d2
240 ≤ 300(0.5)2
240 ≤ 75
Since the chosen point is not a solution,
shade the region outside (below) the
parabola.
The point (0.75,60) lies in the shaded region
of the graph from part (a), so (0.75,60) is a
solution of the given inequality.
Therefore, a shelf that is 0.75 inch thick can
support a weight of 60 pounds.
W
300
250
200
150
100
50
0
0.5
y
1.0
1.5
d
GRAPHING A SYSTEM OF
Graph the system of quadratic inequalities.
y≥x2-4
Inequality 1
y&lt;-x2-x+2 Inequality 2
Solution:
1. Graph the inequality y≥x2-4. The graph is
in red region inside and including the
parabola y = x2-4.
2. Graph the inequality y&lt;-x2-x+2. The
graph is in blue region inside (but not
including) the parabola y = -x2-x+2.
3. Identify the region where two graphs
overlap. This region is the graph of the
system.
y
y≥x2-4
x
y&lt;-x2-x+2.
ONE VARIABLE
1. To solve ax2 + bx + c &lt; 0 (or ax2 + bx + c ≤ 0),
graph y = ax2 + bx + c and identify the x values
for which the graph lies below (or on and
below)
the
x-axis.
2. To solve ax2 + bx + c &gt; 0 (or ax2 + bx + c ≥ 0),
graph y = ax2 + bx + c and identify the x values
for which the graph lies above (or on and
below) the x-axis.
EXAMPLE
Solve x2 - 6x + 5 &lt; 0
Solution:
1. The solution consist of the x-values for
which the graph of y = x2 - 6x + 5 lies
below the x-axis.
2. Find the graph’s x-intercepts by letting y
=
0
0
=
x2
6x
+
5
0
=
(x-1)(x-5)
x
=
1
or
x
=
5
Sketch a parabola that opens up and has
1 and 5 as x-intercepts.
3. The graph lies below the x-axis between
x = 1 and x = 5.
4. The solution of the given inequality is
1&lt;x&lt;5.
y
1
1
3
5
x
EXAMPLE
Solve 2x2 + 3x -3 ≥ 0
Solution:
1. The solution consist of the x-values for which the graph of
y = 2x2 + 3x -3 lies on and above the x-axis.
2. Find the graph’s x-intercepts by letting y = 0
0 = 2x2 + 3x -3
x
3 
32  4(2)(3)
2(2)
3  33
4
x  0.69or
x
x  2.19
EXAMPLE(contd.)
y
1. Sketch a parabola that opens up
and has 0.69 and -2.19 as xintercepts.
1
2. The graph lies on and above the xaxis to the left of ( and including) x
= -2.19 and to the right of ( and
including) x = 0.69.
3. The solution of the given inequality
is approximately x ≤ -2.19 or x ≥
0.69.
-2.19
0.69
4
x
INEQUALITY ALGEBRAICALLY
Solve:
x2 + 2x ≤ 8
Solution:
First replace the inequality symbol with equal sign.
x2 + 2x = 8
x2 + 2x - 8 = 0
(x+4)(x-2) = 0
x = - 4 or x = 2
The numbers -4 and 2 are the critical x-values of the inequality x2 + 2x
≤ 8. Plot -4 and 2 on a number line.
SOLUTION (contd.)
-6 -5 -4 -3 -2 -1 0 1 2 3
4 5 6
These critical x-values partition the number line into
three intervals.
Test x = -5, (-5)2 + 2(-5) = 15 ≤ 8
Test x = 0, (0)2 +2(0) = 0 ≤ 8
Test x = -3, (3)2 + 2(3) = 15 ≤ 8
Hence the solution is -4 ≤ x ≤ 2.
REAL LIFE EXAMPLE
For a driver aged x years, a study found that the driver’s reaction time V(x)
(in milliseconds) to a visual stimulus such as traffic can be modeled by:
V(x) = 0.005x2 - 0.23x + 22,
16 ≤ x ≤ 70
At what ages does a driver’s reaction time tend to be greater than 25
milliseconds?
Solution:
The values of x for which
V(x) &gt; 25
0.005x2 - 0.23x +22 &gt; 25
0.005x2 - 0.23x - 3 &gt; 0
The solution consists of the x- values for which the graph lies above the xaxis.
SOLUTION(contd.)
The graph’s x intercept is found by letting y =0 and using the quadratic
formula to solve for x.
0.005x2 - 0.23x - 3 = 0
0.23 (0.23)2  4  0.005  (3)
x
2  0.005
0.23 .0529  .06
x
.01
0.23 .1129
x
.01
0.23 0.34
x
.01
x  57approx.
or
x  11approx

Rejecting the negative value, the
graph’s x-intercept is about 57. The
graph of 0.005x2 - 0.23x - 3 = 0 lies in
the domain 16 ≤ x ≤ 70. The graph lies
above the x-axis when 57 &lt; x ≤ 70.
Hence the drivers over 57 years old
tend to have reaction times greater than
25 milliseconds.
MODELING WITH
VERTEX FORM
Write the quadratic function for the parabola
shown.
Solution:
The vertex shown is (h,k) = (2,-3)
Using the vertex form of the quadratic function.
y = a(x-h)2 + k
y = a(x-2)2 - 3
Use the other given point (4,1) to find a.
1 = a(4-2)2 - 3
1 = 4a - 3
4 = 4a
1=a
Hence the quadratic function for the parabola is
y = (x-2)2 -3
y
1
(4,1)
x
1
(2,-3)
INTERCEPT FORM
Write the quadratic function for the parabola
shown.
Solution:
The x intercepts shown are p = -2, q = 3
Using the intercept form of the quadratic
function.
y = a(x-p)(x-q)
y = a(x+2)(x-3)
Use the other given point (-1,2) to find a.
2 = a(-1+2)(-1-3)
2 = -4a
-1/2 = a
Hence the quadratic function for the parabola
is y = -1/2(x+2)(x-3)
(-1,2)
1
(-2)
(3)
1
x
y
END