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McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
3. Simplifying Radical Expressions
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Exercise 59 Page 250

Recall the formula for factoring a perfect square trinomial.

H

Practice makes perfect

To solve the given quadratic equation, we will start by paying close attention to its terms. Note that the first and last terms are perfect squares, and that the middle term is twice the product of their roots. Therefore, it is a perfect square trinomial and we can factor it following the corresponding formula. a^2-2ab+b^2=(a-b)^2Let's solve the equation by factoring the perfect square trinomial on the left-hand side.

a^2-2a+1=25
WritePow

Write as a power

a^2-2a+1^2=25
SplitIntoFactors

Split into factors

a^2-2(a)(1)+1^2=25
FacNegPerfectSquare

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

(a-1)^2=25
SqrtEqn

sqrt(LHS)=sqrt(RHS)

a-1=± 5
AddEqn

LHS+1=RHS+1

a=± 5+1

We will now find the first and second solutions by using the positive and negative signs.

a=± 5+1
a=5+1 a=- 5+1
a=6 a=- 4

The solutions to our equations are a_1=6 and a_2=- 4. This corresponds to answer H.

Checking Our Answer

 Checking our answer
We can check our answers by substituting them for a in the given equation. Let's start with a=6.

a^2-2a+1=25
Substitute

a= 6

6^2-2( 6)+1? =25
CalcPow

Calculate power

36-2(6)+1? =25
MultNegPos

(- a)b = - ab

36-12+1? =25
AddSubTerms

Add and subtract terms

25=25 ✓

Since substituting and simplifying created a true statement, we know that a=6 is a solution of the equation. Let's check if a=- 4 is also a solution.

a^2-2a+1=25
Substitute

a= - 4

- 4^2-2( - 4)+1? =25
CalcPow

Calculate power

16-2(- 4)+1? =25
MultNegNegOnePar

- a(- b)=a* b

16+8+1? =25
AddTerms

Add terms

25=25 ✓

We created another true statement, so we know that a=- 4 is also a solution of the equation.

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Exercises 2 (Page 248)
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