SECTION 6.5 Solving Quadratic Equations by Factoring^395
CAUTION A common error is to include the common factor 2 as a solution in
Example 3.Only factors containing variables lead to solutions,such as the factor
yin the equation y 13 y- 42 = 0 in Example 1(b).
NOW TRY
EXERCISE 4
Solve each equation.
(a)
(b)
(c)p 16 p- 12 = 2
m^2 = 5 m9 x^2 - 64 = 0Standard formSolving Quadratic EquationsSolve each equation.
(a)
Zero-factor property
Solve each equation.m=-
5
4
or m=
5
4
4 m=-5 or 4 m= 5
4 m+ 5 =0 or 4 m- 5 = 0
14 m+ 5214 m- 52 = 0
16 m^2 - 25 = 0
EXAMPLE 4
Factor the difference of
squares. (Section 6.4)Checkthe solutions, and in the original equation. The solution set is
(b)
Standard form
Factor.
Zero-factor property
Solve.The solution set is 5 0, 2 6.
y= 2
y=0ory- 2 = 0
y 1 y- 22 = 0
y^2 - 2 y= 0
y^2 = 2 y
E-
54 ,
54 F.
5- 4 ,
5
4(c)
Distributive property
Subtract 3.
Factor.
Zero-factor property
Solve each equation.The solution set is E- 32 , 1F. NOW TRY
k=-
3
2
2 k=- 3 k= 1
2 k+ 3 =0ork- 1 = 0
12 k+ 321 k- 12 = 0
2 k^2 +k- 3 = 0
2 k^2 +k= 3
k 12 k+ 12 = 3
Don’t forget to set
the variable factor y
equal to 0.To be in standard
form, 0 must be on
the right side.CAUTION In Example 4(b),it is tempting to begin by dividing both sides of
by yto get Note, however, that we do not get the other solution, 0, if we
divide by a variable. (We maydivide each side of an equation by a nonzeroreal
number, however. For instance, in Example 3we divided each side by 2.)
In Example 4(c),we could not use the zero-factor property to solve the equation
in its given form because of the 3 on the right. The zero-factor property applies only
to a product that equals 0.
k 12 k+ 12 = 3
y=2.
y^2 = 2 y
NOW TRY ANSWERS
- (a) (b)
(c) E-^12 ,^23 FE- 38 , 38 F 5 0, 5 6