Exercise 17 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. r^2 -7r +6 = 0 ⇕ 1r^2+( - 7)r+ 6=0 We can see that a= 1, b= - 7, and c= 6. Let's substitute these values into the Quadratic Formula.

r=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

r=- ( -7)±sqrt(( - 7)^2-4( 1)( 6))/2( 1)

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Solve for r and Simplify

NegNeg

- (- a)=a

r=7±sqrt((- 7)^2-4(1)(6))/2(1)

NegBaseToPosBase

(- a)^2=a^2

r=7±sqrt(49-4(1)(6))/2(1)

Multiply

Multiply

r=7±sqrt(49-24)/2

SubTerm

Subtract term

r=7±sqrt(25)/2

CalcRoot

Calculate root

r=7± 5/2

Using the Quadratic Formula, we found that the solutions of the given equation are r= 7± 52.

r=7± 5/2
r_1=7+5/2	r_2=7-5/2
r_1=12/2	r_2=2/2
r_1= 6	r_2= 1

Therefore, the solutions are r_1= 6 and r_2= 1. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check r_1=6 and r_2=1 one at a time.

r_1=6

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

sqrt(r+3)=r-3

Substitute

r= 6

sqrt(6+3)? = 6-3

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Simplify

AddSubTerms

Add and subtract terms

sqrt(9)? = 3

CalcRoot

Calculate root

3=3 ✓

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

r_2=1

Now, let's substitute r= 1.

sqrt(r+3)=r-3

Substitute

r= 1

sqrt(1+3)? = 1-3

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Simplify

AddSubTerms

Add and subtract terms

sqrt(4)? = -2

CalcRoot

Calculate root

2 ≠ -2 *

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