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McGraw Hill Integrated II, 2012

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McGraw Hill Integrated II, 2012 View details

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Exercise 12 Page 803

Consider the positive and negative solutions when isolating a variable raised to the power of two.

± sqrt(111)

Practice makes perfect

To solve the given equation by taking the square roots, we need to consider the positive and negative solutions.

a^2+6^2=( 7sqrt(3))^2

Calculate power

a^2+36=49* 3

Multiply

a^2+36=147

LHS-36=RHS-36

a^2=111

sqrt(LHS)=sqrt(RHS)

sqrt(a^2)=sqrt(111)

sqrt(a^2)=± a

a=± sqrt(111)

We found that a=± sqrt(111). Therefore, there are two solutions for the equation, which are a=sqrt(111) and a=- sqrt(111).

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=- sqrt(111).

a^2+6^2=( 7sqrt(3))^2

a= - sqrt(111)

( - sqrt(111))^2+6^2? =( 7sqrt(3))^2

▼

Simplify

NegBaseToPosPow

(- a)^2 = a^2

( sqrt(111))^2+6^2? =( 7sqrt(3))^2

( sqrt(a) )^2 = a

111+6^2? =( 7sqrt(3))^2

Calculate power

111+36? =49* 3

Multiply

111+36? =147

Add terms

147=147 ✓

Since 147=147, we know that a=- sqrt(111) is a solution of the equation. Let's check if a=sqrt(111) is also a solution.

a^2+6^2=(7sqrt(3))^2

a= sqrt(111)

( sqrt(111))^2+6^2? =(7sqrt(3))^2

▼

Simplify

( sqrt(a) )^2 = a

111+6^2? =(7sqrt(3))^2

Calculate power

111+36? =49* 3

Multiply

111+36? =147

Add terms

147=147 ✓

Again, since 147=147, we know that a=sqrt(111) is a solution of the equation.

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