Exercise 49 Page 62

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

WritePow

Write as a power

10^2-5^2x^2=0

ProdPowII

a^m b^m = (a b)^m

10^2-(5x)^2=0

FacDiffSquares

a^2-b^2=(a+b)(a-b)

(10+5x)(10-5x)=0

Now we are ready to use the Zero Product Property.

(10+5x)(10-5x)=0

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Solve using the Zero Product Property

ZeroProdProp

Use the Zero Product Property

lc10+5x=0 & (I) 10-5x=0 & (II)

SubEqn

(I): LHS-10=RHS-10

l5x=- 10 10-5x=0

DivEqn

(I): .LHS /5.=.RHS /5.

lx=- 2 10-5x=0

SubEqn

(II): LHS-10=RHS-10

lx=- 2 - 5x=-10

DivEqn

(II): .LHS /- 5.=.RHS /- 5.

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.