McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
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Exercise 11 Page 803

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

± 15

Practice makes perfect
To solve the given equation by taking the square roots, we need to consider the positive and negative solutions.
8^2+b^2=17^2
64+b^2=289
b^2=225
sqrt(b^2)=sqrt(225)

sqrt(a^2)=± a

b=±sqrt(225)
b=± 15
We found that b=± 15. Therefore, there are two solutions for the equation, which are b=15 and b=- 15.

Checking Our Answer

Checking our answer
We can check our answers by substituting them for b in the given equation. Let's start with b=- 15.
8^2+b^2=17^2
8^2+( - 15)^2? =17^2
Simplify
8^2+15^2? =17^2
64+225? =289
289=289 ✓
We obtained a true statement, so b=- 15 is a solution of the equation. Let's check if b=15 is also a solution.
8^2+b^2=17^2
8^2+ 15^2? =17^2
Simplify
64+225? =289
289=289 ✓
Again we got a true statement, so we know that b=15 is a solution of the equation.