Exercise 41 Page 266

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

Finding the Solution

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. x^2 -17x +60 = 0 ⇔ 1x^2+( - 17)x+ 60=0 We can see that a= 1, b= - 17, and c= 60. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -17)±sqrt(( - 17)^2-4( 1)( 60))/2( 1)

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Solve for x and Simplify

NegNeg

- (- a)=a

x=17±sqrt((- 17)^2-4(1)(60))/2(1)

CalcPow

Calculate power

x=17±sqrt(289-4(1)(60))/2(1)

Multiply

Multiply

x=17±sqrt(289-240)/2

SubTerm

Subtract term

x=17±sqrt(49)/2

CalcRoot

Calculate root

x=17± 7/2

Using the Quadratic Formula, we found that the solutions of the given equation are x= 17± 7 2. Therefore, the solutions are x_1=5 and x_2=12. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check x_1=5 and x_2=12 one at a time.

x_1=5

Let's substitute x=5.

sqrt(x+4)=x-8

Substitute

x= 5

sqrt(5+4)? =5-8

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Simplify

AddSubTerms

Add and subtract terms

sqrt(9)? = -3

CalcRoot

Calculate root

3 ≠ -3 *

We got a false statement, so x=5 is an extraneous solution.

x_2=12

Now, let's substitute x=12 into the original equation.

sqrt(x+4)=x-8

Substitute

x= 12

sqrt(12+4)? =12-8

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Simplify

AddSubTerms

Add and subtract terms

sqrt(16)? = 4

CalcRoot

Calculate root

4=4 ✓

In this case we got a true statement. Therefore, x=12 is the only solution of the original equation.