Exercise 52 Page 648

To solve equations with a variable expression inside a radical, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation using this method!

sqrt(5-4x)-6=7

AddEqn

LHS+6=RHS+6

sqrt(5-4x)=13

RaiseEqn

LHS^2=RHS^2

(sqrt(5-4x))^2=13^2

PowSqrt

( sqrt(a) )^2 = a

5-4x=13^2

CalcPow

Calculate power

5-4x=169

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Solve for x

SubEqn

LHS-5=RHS-5

- 4x=164

DivEqn

.LHS /(- 4).=.RHS /(- 4).

x=- 41

Finally, we will check our solution by substituting the value into the original equation.

sqrt(5-4x)-6=7

Substitute

x= - 41

sqrt(5-4( - 41))-6? =7

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Simplify left-hand side

MultNegNegOnePar

- a(- b)=a* b

sqrt(5+164)-6? =7

AddTerms

Add terms

sqrt(169)-6? =7

CalcRoot

Calculate root

13-6? =7

SubTerm

Subtract term

7=7 ✓

Therefore, we know that x=- 41 is a solution of our equation. This corresponds to answer G.