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McGraw Hill Integrated II, 2012

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McGraw Hill Integrated II, 2012 View details

Study Guide and Review

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Exercise 69 Page 84

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.

- 5, 5

Practice makes perfect

To solve the given equation by factoring, we will start by identifying the values of a, b, and c. a^2-25=0 ⇔ 1a^2+ 0a+( - 25)=0 Notice that this equation follows a special pattern. It can be factored as a difference of squares. Let's factor the equation!

a^2-25=0

Write as a power

a^2-5^2=0

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

(a+5)(a-5)=0

Now we are ready to use the Zero Product Property.

(a+5)(a-5)=0

▼

Solve using the Zero Product Property

Use the Zero Product Property

lca+5=0 & (I) a-5=0 & (II)

(I): LHS-5=RHS-5

la=- 5 a-5=0

(II): LHS+5=RHS+5

la=- 5 a=5

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

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

Subchapter links

1 Adding and Subtracting Polynomials p.81

2 Multiplying a Polynomial by a Monomial p.81

3 Multiplying Polynomials p.81

4 Special Products p.82

5 Using the Distributive Property p.82

6 Solving x^2+bx+c=0 p.83

7 Solving ax^2+bx+c=0 p.83

8 Differences of Squares p.84

9 Perfect Squares p.84

10 Roots and Zeros p.85