Exercise 23 Page 494

To solve equations with a variable expression inside a radical, 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-x-2=0 ⇔ ( 1)x^2+( - 1)x+( - 2)=0We can see that a= 1, b= - 1, and c= - 2. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

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

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=1±sqrt((- 1)^2-4(1)(- 2))/2(1)

CalcPow

Calculate power

x=1±sqrt(1-4(1)(- 2))/2(1)

IdPropMult

Identity Property of Multiplication

x=1±sqrt(1-4(- 2))/2

MultNegNegOnePar

- a(- b)=a* b

x=1±sqrt(1+8)/2

AddTerms

Add terms

x=1±sqrt(9)/2

CalcRoot

Calculate root

x=1± 3/2

This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.

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

Therefore, the solutions are x_1= 2 and x_2= - 1. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check x_1=2 and x_2=- 1 one at a time.

x_1=2

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

sqrt(x+2)=x

Substitute

x= 2

sqrt(2+2)? = 2

▼

Simplify

AddTerms

Add terms

sqrt(4)? = 2

CalcRoot

Calculate root

1 ≠ 2 ✓

We got a true statement. Therefore, x=2 is the solution of the original equation.

x_2=- 1

Now, let's substitute x= - 1.

sqrt(x+2)=x

Substitute

x= - 1

sqrt(- 1+2)? = - 1

▼

Simplify

AddTerms

Add terms

sqrt(1)? = - 1

CalcRoot

Calculate root

1=- 1 *

In this case we got a false statement, so x=- 1 is an extraneous solution. Therefore, we obtain that x=2 is the only solution of the original equation.