# Solving Radical Equations

*This post about solving radical equations is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.*

## Question

Does the equation have a real solution? If so, what is it?

## Solution

This kind of equation is called a *radical equation*, because it contains a radical — in this case, a square root.

Let’s try to solve this radical equation:

Subtract 5 from each side:

The square root of something equals something negative? Really? The definition of square root specifies that it means a positive number. But this square root is supposed to equal something negative.

**No real solution.**

You may be tempted to keep going from

and see what happens. OK, let’s try that. Follow the usual steps for solving a radical equation:

- Isolate the radical. We did that above when we added 10 to each side.
- Square both sides:

- Subtract 3 from each side:

- Divide by 4:

There’s a solution: . Now let’s see if it works. Substitute back into the left-hand side of the original equation. You should get:

That simplifies to:

So what looked like the wrong way was indeed the wrong way, the solution does not work, and the answer to the question is **No, the equation does not have a real solution**.

This question is similar to question number 14 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

## Leave a Reply

Want to join the discussion?Feel free to contribute!