Exercise 7 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. Since in our case a is the variable, let's use letters d, e and f instead. a^2 -9a +18 = 0 ⇕ 1a^2+( - 9)a+ 18=0 We can see that d= 1, e= - 9, and f= 18. Let's substitute these values into the Quadratic Formula.

a=- e±sqrt(e^2-4df)/2d

SubstituteValues

Substitute values

a=- ( -9)±sqrt(( - 9)^2-4( 1)( 18))/2( 1)

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Solve for a and Simplify

NegNeg

- (- a)=a

a=9±sqrt((- 9)^2-4(1)(18))/2(1)

NegBaseToPosBase

(- a)^2=a^2

a=9±sqrt(81-4(1)(18))/2(1)

Multiply

Multiply

a=9±sqrt(81-72)/2

SubTerm

Subtract term

a=9±sqrt(9)/2

CalcRoot

Calculate root

a=9± 3/2

Using the Quadratic Formula, we found that the solutions of the given equation are a= 9± 32.

a=9± 3/2
a_1=9+3/2	a_2=9-3/2
a_1=12/2	a_2=6/2
a_1= 6	a_2= 3

Therefore, the solutions are a_1= 6 and a_2= 3. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check a_1=6 and a_2=3 one at a time.

a_1=6

Let's substitute a= 6 into the original equation.

sqrt(a-2)+4=a

Substitute

a= 6

sqrt(6-2)+4? = 6

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Simplify

SubTerms

Subtract terms

sqrt(4)+4? =6

CalcRoot

Calculate root

2+4? =6

AddTerms

Add terms

6=6 ✓

We got a true statement, so a=6 is a solution.

a_2=3

Now, let's substitute a= 3.

sqrt(a-2)+4=a

Substitute

a= 3

sqrt(3-2)+4? = 3

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Simplify

SubTerms

Subtract terms

sqrt(1)+4? =3

CalcRoot

Calculate root

1+4? =3

AddTerms

Add terms

5 ≠ 3 *

In this case we got a false statement. Therefore, a=6 is the only solution to the original equation.