Download Solve quadratic equations by factoring.

Chapter 6
Section 5
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6.5
1
2
Solving Quadratic Equations by
Factoring
Solve quadratic equations by factoring.
Solve other equations by factoring.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solving Quadratic Equations by Factoring.
A quadratic equation is an equation that can be written
the form
ax2 + bx + c = 0,
where a, b, and c are real numbers, with a ≠ 0.
in
The form ax2 + bx + c = 0 is the standard form of a
quadratic equation. For example,
x 2  5x  6  0,
2 x 2  5x  3,
x2  4
and
are all quadratic equations, but only x2 + 5x +6 = 0 is in
standard form.
Until now, we have factored expressions, including many
quadratic expressions. In this section we see how we can use
factored quadratic expressions to solve quadratic equations.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 3
Objective 1
Solve quadratic equations by
factoring.
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Slide 6.5 - 4
Solve quadratic equations by factoring.
We use the zero-factor property to solve a
quadratic equation by factoring.
If a and b are real numbers and if ab = 0,
then a = 0 or b = 0.
That is, if the product of two numbers is 0, then at
least one of the numbers must be 0. One number must,
but both may be 0.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 5
EXAMPLE 1
Solve.
Using the Zero-Factor Property
Solution:
 2x  35x  7   0
 3 7
 ,  
 2 5
x  2 x  4  0
0, 2
2 x  3  0 or 5x  7  0
2x  3  3  0  3 5x  7  7  0  7
2 x 3
5 x 7


2
2
5
5
3
7
x
x
2
5
x0
or
2x  4  0
2x  4  4  0  4
2 x 4

2
2
x  2
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 6
EXAMPLE 2
Solve.
Solving Quadratic Equations
Solution:
x2  2 x  8
x2  2x  8  8  8
x2  2x  8  0
 x  4 x  2  0
x40
x2  0
or
x44  04 x 22  02
x2
x  4
4, 2
or
x5  0
x 2  x  30
x6  0
x 55  05
x2   x  30  x  30   x  30 x  6  6  0  6
x6
x  5
x 2  x  30  0
 x  6 x  5  0
5,6
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 7
Solve quadratic equations by factoring. (cont’d)
In summary, follow these steps to solve quadratic equations by
factoring.
Step 1: Write the equation in standard form— that is, with
all terms on one side of the equals sign in descending
power of the variable and 0 on the other side.
Step 2: Factor completely.
Step 3: Use the zero-factor property to set each factor with
variable equal to 0, and solve the resulting equations.
Step 4: Check each solution in the original equation.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 8
EXAMPLE 3
Solving a Quadratic Equation
with a Common Factor
m2  0
m 2  2  0  2
Solution:
m  2
3m2  9m  30  30  30
2
3m  9m  30  0
m5  0
2
3  m  3m  10   0
m55  05
3  m  2 m  5  0
m5
2,5
Solve 3m2 − 9m = 30.
A common error is to include the common factor 3 as a solution.
Only factors containing variables lead to solutions.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 9
EXAMPLE 4
Solving Quadratic Equations
Solve.
49 x 2  9  0 7 x  3  3  0  3
Solution:
 7 x  3 7 x  3  0
 3 3
 , 
 7 7
x 2  3x
x 2  3x  3 x  3 x
x  x  3  0
7x  3  3  0  3
7x 3

7
7
3
x
7
7 x 3

7
7
x0
x 33  0  3
x
0,3
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
3
7
x 3
Slide 6.5 - 10
EXAMPLE 4
Solving Quadratic Equations
(cont’d)
x20
x22  02
x  2
Solve.
x  4x  7  2
Solution:
4x  7x  2  2  2
2
4x  7x  2  0
 x  2 4x 1  0
2
4 x 1  0
4 x 1  1  0  1
1

 2, 
4

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
4x 1

4
4
1
x
4
Slide 6.5 - 11
Objective 2
Solve other equations by factoring.
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Slide 6.5 - 12
EXAMPLE 5
Solving Equations with More
than Two Variable Factors
2 x3  50 x  0
Solve.
Solution: 2 x x 2  25  0


2 x  x  5 x  5  0
2x  0
2x 0

2 2
x0
x 55  05
x  5
x 55  05
x 5
0, 5,5
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 13
EXAMPLE 5
Solve.
Solving Equations with More
than Two Variable Factors
(cont’d)
 2 x  1  2 x 2  7 x  15  0
Solution:  2 x  1 2 x  3 x  5  0
2 x  1 1  0 1
2 x 1

2
2
1
x
2
2x  3  3  0  3
2x 3

2 2
3
x
2
x 55  05
x  5
1 3

5,  , 
2 2

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 14
EXAMPLE 6
Solve.
Solving an Equation Requiring
Multiplication before Factoring
 x  1 2 x  1   x  1
2
2 x 2  3x  1  x 2  2 x  1
2
2
2
2
2 x  3x  1   x  2 x  1  x  2 x  1   x  2 x  1
x2  5x  0
x  x  5  0
Solution:
x0
x 55  05
x 5
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
 0,5
Slide 6.5 - 15