McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
Study Guide and Review
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Exercise 39 Page 266

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

32

Practice makes perfect

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

Finding the Solution

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. In this case, we will raise both sides of the equation to the second power. Let's do it!
sqrt(a+4)=6
(sqrt(a+4))^2=6^2
a+4=6^2
a+4=36
a=32
The solution of our equation is a= 32. Now, let's check whether our solution is extraneous.

Checking the Solution

To check our solution, we will substitute 32 for a into the original equation. If we obtain a true statement, the solution is not extraneous. Otherwise, the solution is extraneous.
sqrt(a+4)=6
sqrt(32+4)? =6
sqrt(36)? =6
6=6 âś“
We obtained a true statement, so a=32 is a solution to the equation.