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EXAMPLE 1 Solve a quadratic equation by finding square roots
Solve x2 – 8x + 16 = 25.
x2 – 8x + 16 = 25 Write original equation.
(x – 4)2 = 25 Write left side as a binomial squared.
x – 4 = +5 Take square roots of each side.
x = 4 + 5 Solve for x.
The solutions are 4 + 5 = 9 and 4 –5 = – 1.
ANSWER
EXAMPLE 2 Make a perfect square trinomial
Find the value of c that makes x2 + 16x + c a perfect square trinomial. Then write the expression as the square of a binomial.
SOLUTION
STEP 1
Find half the coefficient of x.
STEP 2
162 = 8
Square the result of Step 1. 82 = 64
STEP 3
Replace c with the result of Step 2. x2 + 16x + 64
EXAMPLE 2 Make a perfect square trinomial
The trinomial x2 + 16x + c is a perfect square when c = 64. Then x2 + 16x + 64 = (x + 8)(x + 8) = (x + 8)2.
ANSWER
GUIDED PRACTICE for Examples 1 and 2
1. x2 + 6x + 9 = 36.
x2 + 6x + 9 = 36 Write original equation.
(x + 3)2 = 36 Write left side as a binomial squared.
x + 3 = + 6 Take square roots of each side.
x = – 3 + 6 Solve for x.
The solutions are – 3 + 6 = 3 and – 3 – 6 = – 9.
ANSWER
Solve the equation by finding square roots.
GUIDED PRACTICE for Examples 1 and 2
2. x2 – 10x + 25 = 1.
x2 – 10x + 25 = 1 Write original equation.
(x – 5)2 = 1 Write left side as a binomial squared.
x – 5 = + 1 Take square roots of each side.
x = 5 + 1 Solve for x.
The solutions are 5 – 1 = 4 and 5 + 1 = 6.
ANSWER
GUIDED PRACTICE for Examples 1 and 2
3. x2 – 24x + 144 = 100.
x2 – 24x + 144 = 100 Write original equation.
(x – 12)2 = 100 Write left side as a binomial squared.
x – 12 = + 10 Take square roots of each side.
x = 12 + 10 Solve for x.
The solutions are 12 – 10 = 2 and 12 + 10 = 22.
ANSWER
GUIDED PRACTICE for Examples 1 and 2
4.
x2 + 14x + c
x2 7x
7x 49
x
x
7
7
SOLUTION
STEP 1
Find half the coefficient of x.
STEP 2
142 = 7
Square the result of Step 1. 72 = 49
STEP 3
Replace c with the result of Step 2. x2 + 14x + 49
Find the value of c that makes the expression a perfect square trinomial.Then write the expression as the square of a binomial.
GUIDED PRACTICE for Examples 1 and 2
The trinomial x2 + 14x + c is a perfect square when c = 49.
Then x2 + 14x + 49 = (x + 7)(x + 7) = (x + 7)2.
ANSWER
GUIDED PRACTICE for Examples 1 and 2
5.
x2 + 22x + c
x2 11x
11x 121
x
x
11
11
SOLUTION
STEP 1
Find half the coefficient of x.
STEP 2
222 = 22
Square the result of Step 1. 112 = 121
STEP 3
Replace c with the result of Step 2. x2 + 22x + 121
GUIDED PRACTICE for Examples 1 and 2
The trinomial x2 + 22x + c is a perfect square when c = 144.
Then x2 + 22x + 144 = (x + 11)(x + 11) = (x + 11)2.
ANSWER
GUIDED PRACTICE for Examples 1 and 2
6.
x2 – 9x + c
x2x
x
SOLUTION
STEP 1
Find half the coefficient of x.
STEP 2
STEP 3
Replace c with the result of Step 2.
92
– x
92
– x
814
92
–
92
–
Square the result of Step 1.
x2 – 9x + 814
92
92=
92( ) =
814
GUIDED PRACTICE for Examples 1 and 2
ANSWER
The trinomial x2 – 9x + c is a perfect square when c = .
Then x2 – 9x + = (x – )(x – ) = (x – )2.
814 81
492
92
92