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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

SubEqn

LHS-7^2=RHS-7^2

k^2 = 18^2-7^2

FacDiffSquares

a^2-b^2=(a+b)(a-b)

k^2 = (18+7)(18-7)

AddTerms

Add terms

k^2 = 25(18-7)

SubTerm

Subtract term

k^2 = 25(11)

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(k^2)=sqrt(25(11))

sqrt(a^2)=± a

k=±sqrt(25(11))

WritePow

Write as a power

k=±sqrt(5^2* 11)

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

k=±sqrt(5^2)* sqrt(11)

SqrtPowToNumber

sqrt(a^2)=a

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

UseCalc

Use a calculator

k ≈ ± 5(3.316624 ...)

Multiply

k ≈ ± 16.58312 ...

RoundDec

Round to 1 decimal place(s)

k ≈ ± 16.6