Exercise 27 Page 261

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

Finding the Solutions

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! We now have a quadratic equation, and we need to find its roots. In order to do that, we can factor the obtained equation. Notice that in our case we have a difference of squares. Let's identify the squares and factor the expression!

x^2 -9=0

WritePow

Write as a power

x^2-3^2=0

FacDiffSquares

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

(x+3)(x-3)=0

Now, we can use the Zero Product Property to determine the roots of our equation.

(x+3)(x-3)=0

ZeroProdProp

Use the Zero Product Property

lcx+3=0 & (I) x-3=0 & (II)

SubEqn

(I): LHS-3=RHS-3

lx=-3 x-3=0

AddEqn

(II): LHS+3=RHS+3

lx=-3 x=3

Therefore, the solutions are x_1= -3 and x_2= 3. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check x_1=-3 and x_2=3 one at a time.

x_1=-3

Let's substitute x= -3 into the original equation.

sqrt(5x^2-9)=2x

Substitute

x= -3

sqrt(5( -3)^2-9)? =2( -3)

▼

Simplify

NegBaseToPosBase

(- a)^2=a^2

sqrt(5(9)-9)? =2(-3)

Multiply

Multiply

sqrt(45-9)? =2(-3)

SubTerms

Subtract terms

sqrt(36)? =2(-3)

CalcRoot

Calculate root

6? =2(-3)

MultPosNeg

a(- b)=- a * b

6 ≠ -6 *

We got a false statement, so x=-3 is not a solution.

x_2=3

Now, let's substitute x= 3.

sqrt(5x^2-9)=2x

Substitute

x= 3

sqrt(5( 3)^2-9)? =2( 3)

▼

Simplify

CalcPow

Calculate power

sqrt(5(9)-9)? =2(3)

Multiply

Multiply

sqrt(45-9)? =6

SubTerms

Subtract terms

sqrt(36)? =6

CalcRoot

Calculate root

6 = 6 ✓

In this case we got a true statement. Therefore, x=3 is the solution to the original equation.