Exercise 5 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. x^2 -13x +30 = 0 ⇕ 1x^2+( - 13)x+ 30=0 We can see that a= 1, b= - 13, and c= 30. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -13)±sqrt(( - 13)^2-4( 1)( 30))/2( 1)

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Solve for x and Simplify

NegNeg

- (- a)=a

x=13±sqrt((- 13)^2-4(1)(30))/2(1)

NegBaseToPosBase

(- a)^2=a^2

x=13±sqrt(169-4(1)(30))/2(1)

Multiply

Multiply

x=13±sqrt(169-120)/2

SubTerm

Subtract term

x=13±sqrt(49)/2

CalcRoot

Calculate root

x=13± 7/2

Using the Quadratic Formula, we found that the solutions of the given equation are x= 13± 72.

x=13± 7/2
x_1=13+7/2	x_2=13-7/2
x_1=20/2	x_2=6/2
x_1= 10	x_2= 3

Therefore, the solutions are x_1= 10 and x_2= 3. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check x_1=10 and x_2=3 one at a time.

x_1=10

Let's substitute x= 10 into the original equation.

sqrt(3x-5)=x-5

Substitute

x= 10

sqrt(3( 10)-5)? = 10-5

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Simplify

Multiply

Multiply

sqrt(30-5)? =10-5

SubTerms

Subtract terms

sqrt(25)? = 5

CalcRoot

Calculate root

5=5 ✓

We got a true statement, so x=10 is a solution.

x_2=3

Now, let's substitute x= 3.

sqrt(3x-5)=x-5

Substitute

x= 3

sqrt(3( 3)-5)? = 3-5

Multiply

Multiply

sqrt(9-5)? =3-5

SubTerms

Subtract terms

sqrt(4)? =-2

CalcRoot

Calculate root

2 ≠ -2 *

In this case we got a false statement. Therefore, x=10 is the only solution to the original equation.