McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
8. Difference of Squares
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Exercise 49 Page 520

Make sure that all of the terms are on the same side of the equation.

- 2, 2

Practice makes perfect

To solve the given equation by factoring, we will start by rewriting it, so that all terms are on the left side of the equality sign. 100=25x^2 ⇔ 100-25x^2=0 Notice that this equation follows a special pattern. It can be factored as a difference of squares. Let's factor the equation!

100-25x^2=0
10^2-5^2x^2=0
10^2-(5x)^2=0
(10+5x)(10-5x)=0

Now we are ready to use the Zero Product Property.

(10+5x)(10-5x)=0
â–¼
Solve using the Zero Product Property
lc10+5x=0 & (I) 10-5x=0 & (II)
l5x=- 10 10-5x=0
lx=- 2 10-5x=0
lx=- 2 - 5x=-10
lx=- 2 x=2

We found that the solutions to the given equation are x=- 2 and x=2. To check our answer, we will graph the related function, y=100-25x^2, using a calculator.

We can see that the x-intercepts are - 2 and 2. Therefore, our solutions are correct.