Exercise 36 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. d^2-d-12=0 ⇕ 1d^2+( - 1)d+( -12)=0We can see that a= 1, b= - 1, and c= -12. Let's substitute these values into the Quadratic Formula.

d=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

d=- ( -1)±sqrt(( - 1)^2-4( 1)( -12))/2( 1)

▼

Evaluate

NegNeg

- (- a)=a

d=1±sqrt((- 1)^2-4(1)(-12))/2(1)

NegBaseToPosBase

(- a)^2=a^2

d=1±sqrt(1-4(1)(-12))/2(1)

IdPropMult

Identity Property of Multiplication

d=1±sqrt(1-4(-12))/2

MultNegNegOnePar

- a(- b)=a* b

d=1±sqrt(1+48)/2

AddTerms

Add terms

d=1±sqrt(49)/2

CalcRoot

Calculate root

d=1± 7/2

Using the Quadratic Formula, we found that the solutions of the given equation are d= 1± 7 2.

d=1± 7/2
d_1=1+7/2	d_2=1-7/2
d_1=8/2	d_2=-6/2
d_1= 4	d_2= -3

Therefore, the solutions are d_1=4 and d_2=-3. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check d_1=4 and d_2=-3 one at a time.

d_1=4

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

sqrt(d+12)=d

Substitute

d= 4

sqrt(4+12)? = 4

▼

Simplify

AddTerms

Add terms

sqrt(16)? =4

CalcRoot

Calculate root

4=4 ✓

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

d_2=-3

Now, let's substitute d= -3.

sqrt(d+12)=d

Substitute

d= -3

sqrt(-3+12)? = -3

▼

Simplify

AddTerms

Add terms

sqrt(9)? =-3

CalcRoot

Calculate root

3 ≠ -3 *

We got a false statement, so d=-3 is an extraneous solution. Therefore, d=4 is the only solution of the original equation.