Exercise 34 Page 262

A radical equation is an equation that contains a radical with a variable in it. We want to write a radical equation with a variable on each side. Let's write an equation that meets these conditions. 5sqrt(2x^2-1)=7x Notice that there are infinitely many ways to write such equation. Therefore, this equation is only one of infinitely many possible solutions. Let's now solve our equation.

Solving the Equation

The first step in solving will be to square both sides of the equation so that we can remove the radical.

5sqrt(2x^2-1)=7x

RaiseEqn

LHS^2=RHS^2

(5sqrt(2x^2-1))^2=(7x)^2

PowProd

(a * b)^m=a^m* b^m

5^2 (sqrt(2x^2-1))^2=7^2x^2

CalcPow

Calculate power

25 (sqrt(2x^2-1))^2=49x^2

PowSqrt

( sqrt(a) )^2 = a

25 (2x^2-1)=49x^2

Now that the radical is gone, we can isolate x using the Properties of Equality.

25 (2x^2-1)=49x^2

Distr

Distribute 25

50x^2 - 25 = 49x^2

SubEqn

LHS-49x^2=RHS-49x^2

x^2-25=0

AddEqn

LHS+25=RHS+25

x^2=25

SqrtEqn

sqrt(LHS)=sqrt(RHS)

x= ± 5

The solutions are x_1=5 and x_2=- 5. We need to check the answers to see if we have any extraneous solutions.

Checking the Solutions

We will check x_1=5 and x_2=- 5 one at a time. Let's start by substituting x= 5 into the original equation.

5sqrt(2x^2-1)=7x

Substitute

x= 5

5sqrt(2( 5)^2-1)? =7( 5)

▼

Simplify

CalcPow

Calculate power

5sqrt(2(25)-1)? =7(5)

Multiply

Multiply

5sqrt(50-1)? =35

SubTerm

Subtract term

5sqrt(49)? =35

CalcRoot

Calculate root

5(7)? =35

Multiply

Multiply

35=35 ✓

We got a true statement, so x=5 is a valid solution. Now, we can substitute x= - 5.

5sqrt(2x^2-1)=7x

Substitute

x= - 5

5sqrt(2( - 5)^2-1)? =7( - 5)

▼

Simplify

NegBaseToPosPow

(- a)^2 = a^2

5sqrt(2(25)-1)? =7(- 5)

Multiply

Multiply

5sqrt(50-1)? =- 35

SubTerm

Subtract term

5sqrt(49)? =- 35

CalcRoot

Calculate root

5(7)? =- 35

Multiply

Multiply

35 ≠ - 35 *

In this case, we got a false statement. Therefore, x=5 is the only solution to the original equation.