Download Lesson 8-6 Quadratic Equations: Perfect Squares

Lesson 8-6 Quadratic Equations: Perfect Squares
Perfect Squares – a polynomial whose factors are the same. (a+b)2 = (a+b) (a+b) = a2 + 2ab + b2
Example 1 Recognize and Factor Perfect Square Trinomials
Determine whether each trinomial is a perfect square trinomial. Write yes or no. If
so, factor it.
a. 25n2 – 10n + 1
1. Is the first term a perfect square?
Yes, 25n2 = (5n)2.
2. Is the last term a perfect square?
Yes, 1 = 12.
3. Is the middle term equal to 2(5n)(1)?
Yes, 10n = 2(5n)(1).
Since all three conditions are satisfied, 25n2 – 10n + 1 is a perfect square trinomial.
25n2 – 10n + 1 = (5n)2 + 2(2x)(3) + (1)2
= (5n – 1)2
2
2
Write as a + 2ab + b .
Factor using the pattern.
b. 4a2 – 8a + 16
1. Is the first term a perfect square?
2. Is the last term a perfect square?
3. Is the middle term equal to 2(2a)(4)?
Yes, 4a2 = (2a)2.
Yes, 16 = 42.
No, 8a  2(2a)(4).
Since the middle term does not satisfy the required condition, 4a2 – 8a + 16 is not a
perfect square trinomial.
Factor Summary
Step 1 Factor out GCF
Step 2 Check for a difference of squares or a perfect square trinomial
Step 3 Apply the factoring patterns for x2+bx+c or ax2+bx+c this includes factor by grouping
Example 2 Factor Completely
Factor each polynomial, if possible. If the polynomial cannot be factored, write prime.
a. 3x2 – 12x + 12
Step 1 The GCF of 3, –12, and 12 is 3, so factor it out.
3x2 – 12x + 12 = 3(x2 – 4x + 4)
Step 2 Factor using the pattern ax2 – bx + c. Are there two numbers with a product of 1(4) or
4 and a sum of –4? Yes, the product of –2 and –2 is 4, and the sum is –4.
3x2 – 12x + 12 = 3(x2 – 4x + 4)
= 3[(x)2 – 2(x)(2) + (2)2]
= 3(x – 2)2
3 is the GCF.
2
2
Write as a – 2ab + b .
a = x and b = 2
b. 2x2 – x – 15
Step 1 The GCF of 2, –1, and –15 is 1.
Step 2 Since 15 is not a perfect square, this is not a perfect square trinomial.
Step 3 Factor using the pattern ax2 + bx + c. Are there two numbers with a product of 2  (-15)
or -30 and a sum of -1? Yes, the product of 5 and -6 is -30 and their sum is -1.
2x2 – x – 15 = 2x2 + mx + px – 15
= 2x2 + 5x + (-6x) – 15
= (2x2 + 5x) + (-6x – 15)
= x(2x + 5) + (-3)(2x + 5)
= (2x + 5)(x – 3)
Write the pattern.
m = 5 and p = -6
Group terms with common factors.
Factor out the GCF from each grouping.
2x + 5 is the common factor.
Example 3 Solve Equations with Repeated Factors
Solve 16x2 + 8x + 1 = 0.
16x2 + 8x + 1 = 0
(4x) + 2(4x)(1) + (1)2 = 0
(4x + 1)2 = 0
(4x + 1)(4x + 1) = 0
4x + 1 = 0
4x = -1
1
x = –4
Original equation
2
Recognize 16x – 8x + 1 as a perfect square trinomial.
Factor the perfect square trinomial.
2
Write (4x + 1) as two factors.
Set the repeated factor equal to zero.
Subtract 1 from each side.
2
Divide each side by 4.
Check the answer in the original equation.
Example 4 Use the Square Root Property
Solve each equation. Check the solutions.
4
a. (x – 2)2 =
9
4
(x – 2)2 = 9
Original equation
x–2=  4
Square Root Property
2
x – 2 = 3
4 2 2
9=33
9
2
x=2 
3
2
2
x = 2 + 3 or x = 2 – 3
8
4
=
=
3
3
4
8
The roots are and .
3
3
b. (x + 1)2 = 10
(x + 1)2 = 10
Separate into two equations.
Simplify.
Check in the original equation.
Original equation
x + 1 =  10
x = –1 
Add 2 to each side.
Square Root Property
10
Subtract 1 from each side.
Since 10 is not a perfect square, the roots are –1 
Using a calculator, –1 +
10 or –1 + 10 and –1 – 10 .
10  2.16 and –1 – 10  –4.16.
Real-World Example 5 Solve an Equation
PHYSICAL SCIENCE A ball is dropped from a height of 35 feet. The formula h = -16t2 + h0
can be used to approximate the number of seconds t it takes for the ball to reach height h from
an initial height h0 in feet. Find the time it takes the ball to reach the ground.
h = -16t2 + h0
0 = -16t2 + 35
-35 = -16t2
2.1875 = t2
1.5  t
Original formula
Replace h with 0 and h0 with 35.
Subtract 35 from each side.
Divide each side by –16.
Take the square root of each side.
Since a negative number does not make sense in this situation, the solution is 1.5 seconds. This means
that it takes about 1.5 seconds for the ball to hit the ground.