Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
4. Adding and Subtracting Rational Expressions
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Exercise 64 Page 689

Make sure that the radical expression is isolated. Then, raise both sides of the equation to the second power.

No solution.

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 need to make sure that the radical expression is isolated. Since the radical is isolated, 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(16y)=-8
(sqrt(16y))^2=(-8)^2
16y=(-8)^2
16y=8^2
16y=64
y=4
The solution of our equation is y= 4. Now, let's check whether our solution is extraneous.

Checking the Solution

To check our solution, we will substitute 4 for y into the original equation. If we obtain a true statement, the solution is not extraneous. Otherwise, the solution is extraneous.
sqrt(16y)=-8
sqrt(16( 4))? =-8
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Simplify
sqrt(64)? =-8
8 ≠ -8 *
We obtained a false statement, so y=4 is an extraneous solution to the equation. Therefore, there is no solution to the equation.