Exercise 32 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. n^2-4n-5=0 ⇕ 1n^2+( - 4)n+( -5)=0We can see that a= 1, b= - 4, and c= -5. Let's substitute these values into the Quadratic Formula.

n=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

n=- ( -4)±sqrt(( - 4)^2-4( 1)( -5))/2( 1)

▼

Evaluate

NegNeg

- (- a)=a

n=4±sqrt((- 4)^2-4(1)(-5))/2(1)

NegBaseToPosBase

(- a)^2=a^2

n=4±sqrt(16-4(1)(-5))/2(1)

IdPropMult

Identity Property of Multiplication

n=4±sqrt(16-4(-5))/2

MultNegNegOnePar

- a(- b)=a* b

n=4±sqrt(16+20)/2

AddTerms

Add terms

n=4±sqrt(36)/2

CalcRoot

Calculate root

n=4± 6/2

Using the Quadratic Formula, we found that the solutions of the given equation are n= 4± 6 2.

n=4± 6/2
n_1=4+6/2	n_2=4-6/2
n_1=10/2	n_2=-2/2
n_1= 5	n_2= -1

Therefore, the solutions are n_1=5 and n_2=-1. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check n_1=5 and n_2=-1 one at a time.

n_1=5

Let's substitute n=5 into the original equation.

n=sqrt(4n+5)

Substitute

n= 5

5? =sqrt(4( 5)+5)

▼

Simplify

Multiply

Multiply

5? =sqrt(20+5)

AddTerms

Add terms

5? =sqrt(25)

CalcRoot

Calculate root

5=5 ✓

In this case we got a true statement. Therefore, n=5 is a solution of the original equation.

n_2=-1

Now, let's substitute n= -1.

n=sqrt(4n+5)

Substitute

n= -1

-1? =sqrt(4( -1)+5)

▼

Simplify

MultPosNeg

a(- b)=- a * b

-1? =sqrt(-4+5)

AddTerms

Add terms

-1? =sqrt(1)

CalcRoot

Calculate root

-1 ≠ 1 *

We got a false statement, so n=-1 is an extraneous solution. Therefore, n=5 is the only solution of the original equation.