Exercise 41 Page 646

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

Finding the Solutions

To solve an equation with a variable expression inside a radical, we will first rearrange the radical equation so that the radical expression is isolated. Then, we can raise both sides of the equation to a power equal to the index of the radical. In this case, we will raise both sides of the equation to the second power. Let's do it! 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 -7x +6 = 0 ⇕ 1x^2+( - 7)x+ 6=0 We can see that a= 1, b= - 7, and c= 6. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

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

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Solve for x and Simplify

NegNeg

- (- a)=a

x=7±sqrt((- 7)^2-4(1)(6))/2(1)

NegBaseToPosBase

(- a)^2=a^2

x=7±sqrt(49-4(1)(6))/2(1)

Multiply

Multiply

x=7±sqrt(49-24)/2

SubTerm

Subtract term

x=7±sqrt(25)/2

CalcRoot

Calculate root

x=7± 5/2

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

x=7± 5/2
x_1=7+5/2	x_2=7-5/2
x_1=12/2	x_2=2/2
x_1= 6	x_2= 1

Therefore, the solutions are x_1= 6 and x_2= 1. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check x_1=6 and x_2=1 one at a time.

x_1=6

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

sqrt(x+3)-1=x-4

Substitute

x= 6

sqrt(6+3)-1? = 6-4

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Simplify

AddSubTerms

Add and subtract terms

sqrt(9)-1? = 2

CalcRoot

Calculate root

3-1? =2

SubTerm

Subtract term

2=2 ✓

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

x_2=1

Now, let's substitute x= 1.

sqrt(x+3)-1=x-4

Substitute

x= 1

sqrt(1+3)-1? = 1-4

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Simplify

AddSubTerms

Add and subtract terms

sqrt(4)-1? = -3

CalcRoot

Calculate root

2-1? =-3

SubTerm

Subtract term

1 ≠ -3 *

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