Exercise 69 Page 644

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. s^2-2s-8=0 ⇔ 1s^2+( - 2)s+( -8)=0We can see that a= 1, b= - 2, and c= -8. Let's substitute these values into the Quadratic Formula.

s=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

s=- ( -2)±sqrt(( - 2)^2-4( 1)( -8))/2( 1)

▼

Evaluate right-hand side

NegNeg

- (- a)=a

s=2±sqrt((- 2)^2-4(1)(-8))/2(1)

NegBaseToPosBase

(- a)^2=a^2

s=2±sqrt(4-4(1)(-8))/2(1)

IdPropMult

Identity Property of Multiplication

s=2±sqrt(4-4(-8))/2

MultNegNegOnePar

- a(- b)=a* b

s=2±sqrt(4+32)/2

AddTerms

Add terms

s=2±sqrt(36)/2

CalcRoot

Calculate root

s=2± 6/2

Using the Quadratic Formula, we found that the solutions of the given equation are s= 2± 6 2.

s=2± 6/2
s_1=2+6/2	s_2=2-6/2
s_1=8/2	s_2=-4/2
s_1= 4	s_2= -2

Therefore, the solutions are s_1=4 and s_2=-2. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check s_1=4 and s_2=-2 one at a time.

s_1=4

Let's substitute s=4 into the original equation.

sqrt(2s+8)=s

Substitute

s= 4

sqrt(2( 4)+8)? = 4

▼

Simplify

Multiply

Multiply

sqrt(8+8)? =4

AddTerms

Add terms

sqrt(16)? =4

CalcRoot

Calculate root

4=4 ✓

In this case we got a true statement. Therefore, s=4 is a solution of the original equation.

s_2=-2

Now, let's substitute s= -2.

sqrt(2s+8)=s

Substitute

s= -2

sqrt(2( -2)+8)? = -2

▼

Simplify

MultPosNeg

a(- b)=- a * b

sqrt(-4+8)? =-2

AddTerms

Add terms

sqrt(4)? =-2

CalcRoot

Calculate root

2 ≠ -2 *

We got a false statement, so s=-2 is an extraneous solution. Therefore, s=4 is the only solution of the original equation.