Exercise 63 Page 533

We will find and check the solutions of the given equation.

Finding the Solutions

To solve equations with variable expressions inside the radicals, we can raise both sides of the equation to a power equal to the index of the radicals. Let's try to solve our equation using this method!

sqrt(5x-3) = sqrt(2x+3)

RaiseEqn

LHS^2=RHS^2

(sqrt(5x-3))^2=(sqrt(2x+3))^2

PowSqrt

( sqrt(a) )^2 = a

5x-3 = 2x+3

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Solve for x

SubEqn

LHS-2x=RHS-2x

3x-3 = 3

AddEqn

LHS+3=RHS+3

3x = 6

DivEqn

.LHS /3.=.RHS /3.

x = 6/3

CalcQuot

Calculate quotient

x = 2

Therefore, the solution is x=2. Let's check it to see if we have an extraneous solution.

Checking the Solutions

Let's substitute x=2 into the original equation.

sqrt(5x-3) = sqrt(2x+3)

Substitute

x= 2

sqrt(5( 2)-3) ? = sqrt(2( 2)+3)

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Simplify

Multiply

Multiply

sqrt(10-3) ? = sqrt(4+3)

AddSubTerms

Add and subtract terms

sqrt(7) = sqrt(7) ✓

We obtained a true statement, so x=2 is a solution.