1.

4 Quadratic Equations
Solving a Quadratic Equation
Completing the Square
The Quadratic Formula
Solving for a Specified Variable
The Discriminant

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1.1
Quadratic Equation in One
Variable
An equation that can be written in the
form ax  bx  c  0
2

where a, b, and c are real numbers

with a ≠ 0, is a quadratic equation.
The given form is called standard
form.

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 Second-degree Equation

A quadratic equation is a second-degree

equation.
This is an equation with a squared variable
term and no terms of greater degree.

x 2  25, 4 x 2  4 x  5  0, 3 x 2  4 x  8

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Zero-Factor Property
If a and b are complex numbers with
ab = 0, then a = 0 or b = 0 or both.

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 Example 1 USING THE ZERO-FACTOR
PROPERTY
Solve 6 x 2  7 x  3

Solution:
6x 2  7x  3

6X  7X  3  0
2
Standard form

(3 x  1)(2x  3)  0 Factor.
Zero-factor
3x  1  0 or 2x  3  0 property.

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 Example 1 USING THE ZERO-FACTOR
PROPERTY
Solve 6 x 2  7 x  3

Solution:
3x  1  0 or 2x  3  0 Zero-factor
property.
3x  1 or 2x  3 Solve each
equation.
1 3
x or x
3 2

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Square Root Property

If x2 = k, then

x k or x k

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 Square-Root Property

That is, the solution of

x2  k
Both solutions
are real if k > 0, is
 
and both are
imaginary if k < 0 k,  k
If k = 0, then this
is sometimes
If k < 0, we write or called a double
the solution set
as
i k   k  solution.

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 Example 2 USING THE SQUARE ROOT
PROPERTY
Solve each quadratic equation.
a. x  17
2

Solution:
By the square root property, the solution set
is

 17 

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 Example 2 USING THE SQUARE ROOT
PROPERTY
Solve each quadratic equation.
b. x 2  25
Solution:
Since 1  i ,

the solution set of x2 = − 25

is 5i .

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 Example 2 USING THE SQUARE ROOT
PROPERTY
Solve each quadratic equation.
c. ( x  4)  12
2

Solution:
Use a generalization of the square root
property.
( x  4)2  12
Generalized square
x  4   12 root property.

x  4  12 Add 4.

x 42 3 12  4 3  2 3
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Solving A Quadratic Equation
By Completing The Square
To solve ax2 + bx + c = 0, by completing the square:

Step 1 If a ≠ 1, divide both sides of the equation by a.

Step 2 Rewrite the equation so that the constant term is
alone on one side of the equality symbol.
Step 3 Square half the coefficient of x, and add this square
to both sides of the equation.
Step 4 Factor the resulting trinomial as a perfect square
and combine like terms on the other side.
Step 5 Use the square root property to complete the
solution.

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 Example 3 USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve x2 – 4x –14 = 0 by completing the
square.
Solution
Step 1 This step is not necessary since a = 1.
Step 2 x 2  4 x  14 Add 14 to both
sides.
x  4 x  4  14  4
2
Step 3 2
 1 
 2 (  4 )   4;
add 4 to both sides.
Step 4 ( x  2)  18
2
Factor; combine
terms.
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 Example 3 USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve x2 – 4x –14 = 0 by completing the
square.
Solution

Step 4 ( x  2)  18
2
Factor; combine terms.

Step 5 x  2   18 Square root property.

Take both
roots. x  2  18 Add 2.

x  2  3 2 Simplify the radical.


The solution set is 2  3 2 . 
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 Example 4 USING THE METHOD OF
COMPLETING THE SQUARE a ≠ 1
Solve 9x2 – 12x + 9 = 0 by completing the
square.
Solution
9 x  12x  9  0
2

4
x  x 1 0
2
Divide by 9. (Step 1)
3
4
x  x  1
2
Add – 1. (Step 2)
3
4 4 4 2

x  x   1 
2  1  4 

 2  3   
4
9
; add
4
9
3 9 9
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 Example 4 USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve 9x2 – 12x + 9 = 0 by completing the
square.
Solution
4 4 4  1  4  4 2

x  x   1 
2

 2  3   
9
; add
4
9
3 9 9
2
 2 5
x     Factor, combine
 3 9 terms. (Step 4)

2 5
x   Square root property
3 9
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 Example 4 USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve 9x2 – 12x + 9 = 0 by completing the
square.
Solution 2 5 Square root property
x  
3 9
2 5 a  i a
x  i Quotient rule for
3 3 radicals

2 5
x  i Add ⅔.
3 3
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 Example 4 USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve 9x2 – 12x + 9 = 0 by completing the
square.
Solution
2 5
x  i Add ⅔.
3 3
2 5 
The solution set is   i .
3 3 

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 The Quadratic Formula

The method of completing the square can

be used to solve any quadratic equation. If
we start with the general quadratic equation,
ax2 + bx + c = 0, a ≠ 0, and complete the
square to solve this equation for x in terms
of the constants a, b, and c, the result is a
general formula for solving any quadratic
equation. We assume that a > 0.

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Quadratic Formula

The solutions of the quadratic equation

ax2 + bx + c = 0, where a ≠ 0, are

b  b  4ac
2
x .
2a

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 Caution Notice that the fraction bar
in the quadratic formula extends under
the – b term in the numerator.

b  b  4ac
2
x .
2a

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 Example 5 USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = – 2

Solution:
x  4x  2  0
2 Write in standard
form.

Here a = 1, b = – 4, c = 2

b  b 2  4ac
x Quadratic formula.
2a

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 Example 5 USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = – 2

Solution:
b  b  4ac
2
x Quadratic formula.
2a

(  4)  (  4)2  4(1)(2)

The fraction
2(1)
bar extends
under – b.

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 Example 5 USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = – 2

Solution:
(  4)  (  4)  4(1)(2)
2

The fraction
2(1)
bar extends
4  16  8

under – b.

2
42 2
 16  8  8  4 2  2 2
2
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 Example 5 USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = – 2

Solution:
42 2
 16  8  8  4 2  2 2
2



2 2 2  Factor out 2 in the numerator.
2
Factor first,
then divide.
 2 2 Lowest terms.


The solution set is 2  2 . 
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 Example 6 USING THE QUADRATIC FORMULA
(NONREAL COMPLEX SOLUTIONS)

Solve 2x2 = x – 4.

Solution:
2x 2  x  4  0 Write in standard form.

( 1)  ( 1)  4(2)(4)

2
x Quadratic formula;
2(2) a = 2, b = – 1, c = 4

Use parentheses and

1  1  32 substitute carefully to
 avoid errors.
4

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 Example 6 USING THE QUADRATIC FORMULA
(NONREAL COMPLEX SOLUTIONS)

Solve 2x2 = x – 4.

Solution:
1  1  32

4
1  31
x 1  i
4
1 31 
The solution set is   i .
4 4 
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 Cubic Equation

The equation x3 + 8 = 0 that follows is called

a cubic equation because of the degree 3
term. Some higher-degree equations can
be solved using factoring and the quadratic
formula.

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 Example 7 SOLVING A CUBIC EQUATION

Solve x 3  8  0.
Solution
x 80
3

 x  2  x  2x  4   0
2 Factor as a sum of
cubes.
x  2  0 or x 2  2x  4  0 Zero-factor property

( 2)  ( 2)  4(1)(4)

2
x  2 or x 
2(1)
Quadratic formula; a = 1, b = – 2, c = 4

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 Example 7 SOLVING A CUBIC EQUATION

Solve x 3  8  0.
Solution
2  12
x Simplify.
2
2  2i 3
x Simplify the radical.
2

x

2 1 i 3  Factor out 2 in the
numerator.
2
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 Example 7 SOLVING A CUBIC EQUATION

Solve x 3  8  0.
Solution

x  1 i 3 Lowest terms


The solution set is 2,1  i 3 . 

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