McGraw Hill Glencoe Geometry, 2012
MH
McGraw Hill Glencoe Geometry, 2012 View details
Get Ready for the Chapter
Continue to next subchapter

Exercise 12 Page 835

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
a^2+36=49* 3
a^2+36=147
a^2=111
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
( - sqrt(111))^2+6^2? =( 7sqrt(3))^2
Simplify
( sqrt(111))^2+6^2? =( 7sqrt(3))^2
111+6^2? =( 7sqrt(3))^2
111+36? =49* 3
111+36? =147
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
( sqrt(111))^2+6^2? =(7sqrt(3))^2
Simplify
111+6^2? =(7sqrt(3))^2
111+36? =49* 3
111+36? =147
147=147 ✓
Again, since 147=147, we know that a=sqrt(111) is a solution of the equation.