McGraw Hill Integrated II, 2012
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Exercise 10 Page 803

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

± 9

Practice makes perfect
To solve the given equation by taking the square roots, we need to consider the positive and negative solutions.
a^2+40^2=41^2
a^2+1600=1681
a^2 = 81
sqrt(a^2)=sqrt(81)

sqrt(a^2)=± a

a = ±sqrt(81)
a=± 9
We found that a=± 9. Therefore, there are two solutions for the equation, which are a=9 and a=- 9.

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=- 9.
a^2+40^2=41^2
( - 9)^2+40^2? =41^2
Simplify
9^2+40^2? =41^2
81+1600? =1681
1681=1681 ✓
Since we got a true statement, we know that a=- 9 is a solution of the equation. Let's check if a=9 is also a solution.
a^2+40^2=41^2
9^2+41^2? =41^2
Simplify
81+1600? =1681
1681=1681 ✓
We obtained a true statement, so we know that a=9 is also a solution of the equation.