Exercise 72 Page 84

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. x^2-5=20 ⇔ x^2-25=0 Now we can identify the values of a, b, and c. x^2-25=0 ⇔ 1x^2+ 0x+( -25)=0 Notice that this equation follows a special pattern. It can be factored as a difference of squares. Let's factor the equation!

x^2-25=0

WritePow

Write as a power

x^2-5^2=0

FacDiffSquares

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

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

Now we are ready to use the Zero Product Property.

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

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Solve using the Zero Product Property

ZeroProdProp

Use the Zero Product Property

lcx+5=0 & (I) x-5=0 & (II)

SubEqn

(I): LHS-5=RHS-5

lcx=- 5 & (I) x-5=0 & (II)

AddEqn

(II): LHS+5=RHS+5

lcx=- 5 & (I) x=5 & (II)

We found that the solutions to the given equation are x=- 5 and x=5. To check our answer, we will graph the related function f(x)=x^2-25 using a calculator.

We can see that the x-intercepts are - 5 and 5. Therefore, our solutions are correct.