Exercise 15 Page 261

We will find and check the solutions of the given equation.

Finding the Solutions

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

y=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

y=- 1±sqrt(1^2-4( 1)( -12))/2( 1)

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Solve for y and Simplify

CalcPow

Calculate power

y=-1±sqrt(1-4(1)(-12))/2(1)

IdPropMult

Identity Property of Multiplication

y=-1±sqrt(1-4(-12))/2

MultPosNeg

a(- b)=- a * b

y=-1±sqrt(1-(-48))/2

SubNeg

a-(- b)=a+b

y=-1±sqrt(1+48)/2

AddTerms

Add terms

y=-1±sqrt(49)/2

CalcRoot

Calculate root

y=-1± 7/2

Using the Quadratic Formula, we found that the solutions of the given equation are y= -1± 72.

y=-1± 7/2
y_1=-1+7/2	y_2=-1-7/2
y_1=6/2	y_2=-8/2
y_1= 3	y_2= -4

Therefore, the solutions are y_1= 3 and y_2= -4. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check y_1=3 and y_2=-4 one at a time.

y_1=3

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

y=sqrt(12-y)

Substitute

y= 3

3? =sqrt(12- 3)

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Simplify

SubTerm

Subtract term

3? = sqrt(9)

CalcRoot

Calculate root

3=3 ✓

We got a true statement, so y=3 is a solution.

y_2=-4

Now, let's substitute y= -4.

y=sqrt(12-y)

Substitute

y= -4

-4? =sqrt(12-( -4))

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Simplify

SubNeg

a-(- b)=a+b

-4? = sqrt(12+4)

AddTerms

Add terms

-4? = sqrt(16)

CalcRoot

Calculate root

-4 ≠ 4 *

In this case we got a false statement. Therefore, y=3 is the only solution to the original equation.