Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
Cumulative Standards Review

Exercise 6 Page 659

If the radical expression is isolated, the first step is to raise both sides of the radical equation to the second power.

I

Practice makes perfect

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

Finding the Solution

The radical expression is isolated, so we do not need to rearrange the radical equation. We can start by raising 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!
sqrt(x-8)=12
(sqrt(x-8))^2=12^2
x-8=12^2
x-8=144
x=152
The solution of our equation is x= 152. Now, let's check whether our solution is extraneous.

Checking the Solution

To check our solution, we will substitute 152 for x into the original equation. If we obtain a true statement, the solution is not extraneous. Otherwise, the solution is extraneous.
sqrt(x-8)=12
sqrt(152-8)? =12
â–Ľ
Simplify
sqrt(144)? =12
12=12 âś“
We obtained a true statement, so x=152 is a solution to the equation. This result corresponds to answer I.