Exercise 42 Page 266

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 -15x +50 = 0 ⇔ 1x^2+( - 15)x+ 50=0We can see that a= 1, b= - 15, and c= 50. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -15)±sqrt(( - 15)^2-4( 1)( 50))/2( 1)

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Solve for x and Simplify

NegNeg

- (- a)=a

x=15±sqrt((- 15)^2-4(1)(50))/2(1)

CalcPow

Calculate power

x=15±sqrt(225-4(1)(50))/2(1)

Multiply

Multiply

x=15±sqrt(225-200)/2

SubTerm

Subtract term

x=15±sqrt(25)/2

CalcRoot

Calculate root

x=15± 5/2

Using the Quadratic Formula, we found that the solutions of the given equation are x= 15± 5 2. Therefore, the solutions are x_1=10 and x_2=5. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check x_1=10 and x_2=5 one at a time.

x_1=10

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

sqrt(3x-14)+x=6

Substitute

x= 10

sqrt(3(10)-14)+10? =6

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Simplify

Multiply

Multiply

sqrt(30-14)+10? = 6

SubTerm

Subtract term

sqrt(16)+10? = 6

CalcRoot

Calculate root

4+10? = 6

AddTerms

Add terms

14 ≠ 6 *

We got a false statement, so x=10 is an extraneous solution.

x_2=5

Now, let's substitute x=5.

sqrt(3x-14)+x=6

Substitute

x= 5

sqrt(3(5)-14)+5? =6

▼

Simplify

Multiply

Multiply

sqrt(15-14)+5? = 6

SubTerm

Subtract term

sqrt(1)+5? = 6

CalcRoot

Calculate root

1+5? = 6

AddTerms

Add terms

6=6 ✓

In this case we got a true statement. Therefore, x=5 is the only solution of the original equation.