Exercise 16 Page 644

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. u^2 -u -6 = 0 ⇕ 1u^2+( - 1)u+( -6)=0 We can see that a= 1, b= - 1, and c= -6. Let's substitute these values into the Quadratic Formula.

u=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

u=- ( -1)±sqrt(( - 1)^2-4( 1)( -6))/2( 1)

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Solve for u and Simplify

NegNeg

- (- a)=a

u=1±sqrt((- 1)^2-4(1)(-6))/2(1)

NegBaseToPosBase

(- a)^2=a^2

u=1±sqrt(1-4(1)(-6))/2(1)

IdPropMult

Identity Property of Multiplication

u=1±sqrt(1-4(-6))/2

MultPosNeg

a(- b)=- a * b

u=1±sqrt(1-(-24))/2

SubNeg

a-(- b)=a+b

u=1±sqrt(1+24)/2

AddTerms

Add terms

u=1±sqrt(25)/2

CalcRoot

Calculate root

u=1± 5/2

Using the Quadratic Formula, we found that the solutions of the given equation are u= 1± 52.

u=1± 5/2
u_1=1+5/2	u_2=1-5/2
u_1=6/2	u_2=-4/2
u_1= 3	u_2= -2

Therefore, the solutions are u_1= 3 and u_2= -2. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check u_1=3 and u_2=-2 one at a time.

u_1=3

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

sqrt(u+6)=u

Substitute

u= 3

sqrt(3+6)? = 3

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Simplify

AddTerms

Add terms

sqrt(9)? =3

CalcRoot

Calculate root

3=3 ✓

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

u_2=-2

Now, let's substitute u= -2.

sqrt(u+6)=u

Substitute

u= -2

sqrt(-2+6)? = -2

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Simplify

AddTerms

Add terms

sqrt(4)? = -2

CalcRoot

Calculate root

2 ≠ -2 *

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