McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Radical Equations
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Exercise 14 Page 261

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

11

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. Let's try to solve our equation using this method!
sqrt(k+7)=3sqrt(2)
(sqrt(k+7))^2=(3sqrt(2))^2
k+7=(3sqrt(2))^2
â–Ľ
Simplify right-hand side
k+7=3^2(sqrt(2))^2
k+7=3^2(2)
k+7=9(2)
k+7=18
k=11
The solution of our equation is k= 11. Now, let's check whether our solution is extraneous.

Checking the Solution

To check our solution, we will substitute 11 for k into the original equation. If we obtain a true statement, the solution is not extraneous. Otherwise, the solution is extraneous.
sqrt(k+7)=3sqrt(2)
sqrt(11+7)? =3sqrt(2)
â–Ľ
Simplify
sqrt(18)? =3sqrt(2)
sqrt(9 * 2)? =3sqrt(2)
sqrt(9) * sqrt(2)? =3sqrt(2)
3sqrt(2)=3sqrt(2) âś“
We obtained a true statement, so k=11 is a solution to the equation.