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McGraw Hill Glencoe Algebra 1, 2012

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McGraw Hill Glencoe Algebra 1, 2012 View details

4. Radical Equations

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Exercise 5 Page 644

Raise both sides of the equation to a power equal to the index of the radical.

10

Practice makes perfect

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!

sqrt(3x-5)=x-5

LHS^2=RHS^2

(sqrt(3x-5))^2=(x-5)^2

( sqrt(a) )^2 = a

3x-5 = (x-5)^2

ExpandNegPerfectSquare

(a-b)^2=a^2-2ab+b^2

3x-5 = x^2 -10x + 25

▼

LHS-(3x-5)=RHS-(3x-5)

LHS-3x=RHS-3x

-5 = x^2 -13x + 25

LHS+5=RHS+5

0 = x^2 -13x +30

Rearrange equation

x^2 -13x +30=0

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)

▼

Solve for x and Simplify

- (- 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

x=13±sqrt(169-120)/2

Subtract term

x=13±sqrt(49)/2

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

x= 10

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

▼

Simplify

Multiply

sqrt(30-5)? =10-5

Subtract terms

sqrt(25)? = 5

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

x= 3

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

Multiply

sqrt(9-5)? =3-5

Subtract terms

sqrt(4)? =-2

Calculate root

2 ≠ -2 *

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

Subchapter links

Check Your Understanding p.644

H.O.T. Problems p.645

Standardized Test Practice p.646

Skills Review p.646