McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Radical Equations
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Exercise 24 Page 261

Rearrange the radical equation so that the radical expression is isolated. Then raise both sides of the equation to the second power.

0.2

Practice makes perfect

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

Finding the Solution

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!
6sqrt(5k/4)-3=0
6sqrt(5k/4)=3
sqrt(5k/4)=3/6
sqrt(5k/4)=1/2
(sqrt(5k/4))^2=(1/2)^2
5k/4=(1/2)^2
â–Ľ
Solve for k
5k/4=1^2/2^2
5k/4=1/4
5k=1
k=1/5
k=0.2
The solution of our equation is k= 0.2. Now, let's check whether our solution is extraneous.

Checking the Solution

To check our solution, we will substitute 0.2 for k into the original equation. If we obtain a true statement, the solution is not extraneous. Otherwise, the solution is extraneous.
6sqrt(5k/4)-3=0
6sqrt(5( 0.2)/4)-3? =0
â–Ľ
Simplify
6sqrt(1/4)-3? =0
6(sqrt(1)/sqrt(4))-3? =0
6(1/2)-3? =0
6/2-3? =0
3-3? =0
0=0 âś“
We obtained a true statement, so k=0.2 is a solution to the equation.