Exercise 31 Page 637

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. To do it, let's identify the values of a, b, and c. x^2 -2x -3 = 0 ⇕ 1x^2+( - 2)x+( -3)=0We can see that a= 1, b= - 2, and c= -3. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -2)±sqrt(( - 2)^2-4( 1)( -3))/2( 1)

▼

Evaluate

NegNeg

- (- a)=a

x=2±sqrt((- 2)^2-4(1)(-3))/2(1)

NegBaseToPosBase

(- a)^2=a^2

x=2±sqrt(4-4(1)(-3))/2(1)

IdPropMult

Identity Property of Multiplication

x=2±sqrt(4-4(-3))/2

MultNegNegOnePar

- a(- b)=a* b

x=2±sqrt(4+12)/2

AddTerms

Add terms

x=2±sqrt(16)/2

CalcRoot

Calculate root

x=2± 4/2

Using the Quadratic Formula, we found that the solutions of the given equation are x= 2± 4 2.

x=2± 4/2
x_1=2+4/2	x_2=2-4/2
x_1=6/2	x_2=-2/2
x_1= 3	x_2= -1

Therefore, the solutions are x_1=3 and x_2=-1. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check x_1=3 and x_2=-1 one at a time.

x_1=3

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

x=sqrt(2x+3)

Substitute

x= 3

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

▼

Simplify

Multiply

Multiply

3? =sqrt(6+3)

AddTerms

Add terms

3? =sqrt(9)

CalcRoot

Calculate root

3=3 ✓

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

x_2=-1

Now, let's substitute x= -1.

x=sqrt(2x+3)

Substitute

x= -1

-1? =sqrt(2( -1)+3)

▼

Simplify

MultPosNeg

a(- b)=- a * b

-1? =sqrt(-2+3)

AddTerms

Add terms

-1? =sqrt(1)

CalcRoot

Calculate root

-1 ≠ 1 *

We got a false statement, so x=-1 is an extraneous solution. Therefore, x=3 is the only solution of the original equation.