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Section 3.2
Quadratic Equations,
Functions, Zeros, and
Models
Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.
Objectives

Find zeros of quadratic functions and solve quadratic
equations by using the principle of zero products, by
using the principle of square roots, by completing the
square, and by using the quadratic formula.
 Solve equations that are reducible to quadratic.
 Solve applied problems using quadratic equations.
Quadratic Equations
A quadratic equation is an equation that can be written
in the form
ax2 + bx + c = 0, a  0,
where a, b, and c are real numbers.
A quadratic equation written in this form is said to
be in standard form.
Quadratic Functions
A quadratic function f is a function that can be written
in the form
f (x) = ax2 + bx + c, a  0,
where a, b, and c are real numbers.
The zeros of a quadratic function f (x) = ax2 + bx + c are
the solutions of the associated quadratic equation
ax2 + bx + c = 0. Quadratic functions can have realnumber or imaginary-number zeros and quadratic
equations can have real-number or imaginary-number
solutions.
Equation-Solving Principles
The Principle of Zero Products:
If ab = 0 is true, then
a = 0 or b = 0,
and if a = 0 or b = 0,
then ab = 0.
Equation-Solving Principles
The Principle of Square Roots:
If x2 = k, then
x
k
or
x   k.
Example
Solve 2x2  x = 3.
2x 2  x  3
2x 2  x  3  0
x  12x  3  0
x 1  0
or
2x  3  0
x  1
or
x  1
or
2x  3
3
x
2
Example (cont)
Check: x = – 1
2x 2  x  3
2
2 1  1 ? 3
2 1  1
3
2 1
3
3
3 TRUE
3
The solutions are –1 and .
2
3
Check: x 
2
2x 2  x  3
2
 3
 3
2     ? 3
 2
 2
9 3
2 
4 2
9 3

2 2
6
2
3
3
3
3
3 TRUE
Example
Solve 2x2  10 = 0.
2x  10  0
2
2x  10
x2  5
x  5 or x   5
2
Check:
2x 2  10  0
2
2  5  10 ? 0

The solutions are 5 and  5 .

2  5  10
0
10  10
0
0
0
TRUE
Completing the Square
To solve a quadratic equation by completing the square:
1. Isolate the terms with variables on one side of the
equation and arrange them in descending order.
2. Divide by the coefficient of the squared term if that
coefficient is not 1.
3. Complete the square by finding half the coefficient of
the first-degree term and adding its square on both
sides of the equation.
4. Express one side of the equation as the square of a
binomial.
5. Use the principle of square roots.
6. Solve for the variable.
Example
Solve 2x2  1 = 3x.
2
2x 2  1  3x
2x 2  3x  1  0
2x 2  3x
1
3
1
2
x  x

2
2
3
9
1 9
2
x  x
 
2
16 2 16
3
17

 x   
4
16
3
17
x  
4
4
3
17
x 
4
4
The solutions are 3  17 .
4
Quadratic Formula
The solutions of ax2 + bx + c = 0, a  0, are given by
b  b 2  4ac
x
.
2a
This formula can be used to solve any quadratic
equation.
Example
Solve 3x2 + 2x = 7. Find exact solutions and approximate
solutions rounded to the thousandths.
3x2 + 2x  7 = 0
a = 3, b = 2, c = 7
The exact solutions are:
2  2  4 37 
2
x
2 3

2  4  84  2  88

6
6
2 1  22
2  2 22


6
23
  1 
3
22
1  22
or
3
The approximate solutions are –1.897 and 1.230.
Discriminant
When you apply the quadratic formula to any quadratic
equation, you find the value of b2  4ac, which can be
positive, negative, or zero. This expression is called the
discriminant.
For ax2 + bx + c = 0, where a, b, and c are real numbers:
b2  4ac = 0 One real-number solution;
b2  4ac > 0 Two different real-number solutions;
b2  4ac < 0 Two different imaginary-number
solutions, complex conjugates.
Equations Reducible to Quadratic
Some equations can be treated as quadratic, provided
that we make a suitable substitution.
Example: x4  5x2 + 4 = 0
Knowing that x4 = (x2)2, we can substitute u for x2 and the
resulting equation is then u2  5u + 4 = 0. This equation
can then be solved for u by factoring or using the
quadratic formula. Then the substitution can be reversed
by replacing u with x2, and solving for x. Equations like
this are said to be reducible to quadratic, or quadratic
in form.
Example
Solve: x4  5x2 + 4 = 0.
x4  5x2 + 4 = 0
u2  5u + 4 = 0
(substituting u for x2)
(u  1)(u  4) = 0
u  1 = 0 or u  4 = 0
u = 1 or
u=4
x2 = 1 or
x2 = 4
(substitute x2 for u
x = ±1 or
x = ±2
and solve for x)
The solutions are 1, 1, 2, and 2.
Applications
Some applied problems can be translated to quadratic
equations.
Example
Time of Free Fall. The Petronas Towers in Kuala Lumpur,
Malaysia are 1482 ft tall. How long would it take an object
dropped from the top reach the ground?
Example (continued)
1. Familiarize. The formula s = 16t2 is used to
approximate the distance s, in feet, that an object falls
freely from rest in t seconds.
2. Translate. Substitute 1482 for s in the formula:
1482 = 16t2.
3. Carry out. Use the principle of square roots.
1482  16t 2
1482
t
16
1482
2
t
16
9.624  t
Example (continued)
4. Check. In 9.624 seconds, a dropped object would
travel a distance of 16(9.624)2, or about 1482 ft. The
answer checks.
5. State. It would take about 9.624 sec for an object
dropped from the top of the Petronas Towers to
reach the ground.