Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
9. Proofs Using Coordinate Geometry
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Exercise 40 Page 418

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

16.6 and - 16.6

Practice makes perfect
We will solve the given equation by taking the square roots. Remember to consider the positive and negative solutions.
7^2+k^2 = 18^2
k^2 = 18^2-7^2
k^2 = (18+7)(18-7)
k^2 = 25(18-7)
k^2 = 25(11)
sqrt(k^2)=sqrt(25(11))

sqrt(a^2)=± a

k=±sqrt(25(11))
k=±sqrt(5^2* 11)
k=±sqrt(5^2)* sqrt(11)
k=± 5sqrt(11)
We found that k=± 5 sqrt(11). Thus, there are two solutions for the equation, which are k= 5 sqrt(11) and k= - 5 sqrt(11). Remember that we are asked to round the solutions to the nearest tenth. Therefore, we will use calculator to find the approximated values of the solutions.
k=± 5 sqrt(11)
Use a calculator
k ≈ ± 5(3.316624 ...)
k ≈ ± 16.58312 ...
k ≈ ± 16.6

Checking Our Answer

Checking our answer
We can check our answers by substituting them for k in the given equation. Let's start with k=- 5 sqrt(11).
7^2+k^2=18^2
7^2+( - 5 sqrt(11))^2? =18^2
Simplify
7^2+(5 sqrt(11))^2? =18^2
7^2+5^2 (sqrt(11))^2? =18^2
7^2+5^2(11)? =18^2
49+25(11)? =324
49+275? =324
324=324 ✓
Since 324=324, we know that k=- 5 sqrt(11) is a solution of the equation. Let's check if k=5 sqrt(11) is also a solution.
7^2+k^2=18^2
7^2+( 5 sqrt(11))^2? =18^2
Simplify
7^2+5^2 (sqrt(11))^2? =18^2
7^2+5^2(11)? =18^2
49+25(11)? =324
49+275? =324
324=324 ✓
Again, since 324=324, we know that k=5 sqrt(11) is a solution of the equation.