Pearson Algebra 1 Common Core, 2015
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Pearson Algebra 1 Common Core, 2015 View details
4. Solving Radical Equations
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Exercise 1 Page 636

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

12

Practice makes perfect

We will find and check the solution of the given radical 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!
sqrt(3x)+10=16
sqrt(3x)=6
(sqrt(3x))^2=6^2
3x=6^2
3x=36
x=12
The solution of our equation is x= 12. Now, let's check whether our solution is extraneous.

Checking the Solution

To check our solution, we will substitute 12 for x into the original equation. If we obtain a true statement, the solution is not extraneous. Otherwise, the solution is extraneous.
sqrt(3x)+10=16
sqrt(3( 12))+10? =16
â–Ľ
Simplify
sqrt(36)+10? =16
6+10? =16
16=16 âś“
We obtained a true statement, so x=12 is a solution to the original equation.