McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
9. Perfect Squares
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Exercise 38 Page 69

Start by identifying the values of a, b, and c. Be sure that all of the terms of are on the same side and in the correct order for the standard form of a quadratic function.

1 and -9

Practice makes perfect
To solve the given equation by factoring, we will start by identifying the values of a, b, and c. x^2+8x+16=25 ⇔ 1x^2+ 8x - 9=0 We have a quadratic equation with a= 1, b= 8, and c= - 9. To factor the left-hand side, we need to find a factor pair of 1 * - 9=- 9 whose sum is 8. Since - 9 is a negative number, we will only consider factors with opposite signs — one positive and one negative — so that their product is negative.
Factor Pair Product of Factors Sum of Factors
1 and - 9 1* (- 9) - 9 1+(- 9) - 8
- 1 and 9 - 1* 9 - 9 - 1+9 8
3 and - 3 3* (- 3) - 9 3+(- 3) 0
The integers whose product is - 9 and whose sum is 8 are - 1 and 9. With this information, we can rewrite the linear factor on the left-hand side of the equation, and factor by grouping.
x^2+8x-9=0
x^2-x+9x-9=0
â–Ľ
Factor out x & 9
x(x-1)+9x-9=0
x(x-1)+9(x-1)=0
(x-1)(x+9)
Now we are ready to use the Zero Product Property.
(x-1)(x+9)=0
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Solve using the Zero Product Property
lcx-1=0 & (I) x+9=0 & (II)
lx=1 x+9=0
lx=1 x=-9
We found that the solutions to the given equation are x=1 and x=-9. To check our answer, we will graph the related functions y=x^2+8x-9 and y=(x-1)(x+9) in the same coordinate plane using a graphing calculator.

We see that only one graph appears. This means that both graphs coincide. We can see that the x-intercepts are 1 and -9. Therefore, our solutions are correct. âś“