# Squares and Square Roots

A radical equation is an equation that features a variable contained inside a radicand. At least it won't get wet if it rains.

An example of a radical equation is .

The equation is not a radical equation, because the variable doesn't occur inside the radicand. The 5 and 9 are making it wait outside, apparently, and it's getting drenched.

To solve a radical equation that has a single radical on one side and a number on the other, we square both sides to eliminate the radical and then solve the resulting equation. It'll only get more complicated from here, so enjoy it while it lasts.

### Sample Problem

Solve: .

We square both sides to eliminate the radical. This gives us x + 1 = 49, so x = 48.

Check to make sure this works. Plugging x = 48 into the left-hand side of the equation yields , which is totally 7.

Sometimes we need to rearrange the equation a little so that the radical is all by itself, or "isolated" on one side of the equation. Poor little radical.

### Sample Problem

Solve: .

First we need to add 1 to each side to get the radical all by itself on the left-hand side of the equation. Don't feel badly for him; he's a loner anyway. He'll appreciate the solitude:

.

Then, we square both sides to eliminate the radical:

5 + x = 16.

Solving this equation gives us x = 11.

We'll push our double-checking agenda again here. Plug x = 11 into the left-hand side of the equation:

...which is, in fact, 3. Looks like our train hasn't come off the tracks. The two sides of the equation agree, so our solution is x = 11.

As we mentioned, some radical equations have no solution and are so obvious about it that we don't need to do any work. We have no problem with that. Subtlety is overrated.

### Sample Problem

Solve: .

The radical is already by itself. It even hung a "Do Not Disturb" sign on its door. We square both sides:

2x + 6 = x2 + 6x + 9

Now we rearrange for a nice quadratic equation, being sure we've got zero off to one side:

0 = x2 + 4x + 3

This equation factors as:

0 = (x + 3)(x + 1)

So our solutions are x = -3 and x = -1.

As always, we'll cover our butts by checking to make sure both of these are correct solutions.

When x = -3, the left-hand side of the original equation is:

And the right-hand side of the original equation is:

-3 + 3 = 0

Since the two sides agree, x = -3 is a solution to the original equation.

When x = -1, the left-hand side of the original equation is:

And the right-hand side is:

(-1) + 3 = 2

So x = -1 is also a solution to the original equation. Excellent. Butts officially covered.

Be careful: When solving radical equations, make sure to check your answers. Sometimes they won't actually be solutions to the original equation. You don't want egg on your face. Especially if it's been soft-boiled.

### Sample Problem

Solve:

We square both sides to find:

4x + 5 = x2

Rearrange for a nice-looking quadratic equation. Wow, has this thing been working out?

0 = x2 – 4x – 5

This factors as:

0 = (x – 5)(x + 1)

So our solutions should be x = 5 and x = -1. Now we check to make sure both of these are solutions to the original equation.

When x = 5, the left-hand side of the original equation is , and the right-hand side of the original equation is also 5, so x = 5 is a solution.

When x = -1, the left-hand side of the original equation is .

However, when x = -1 the right-hand side of the original equation is -1. Uh-oh.

Since the left-hand and right-hand sides of the equation don't agree for x = -1, this isn't a solution. Argh, so close: x = -1 is an extraneous solution. In other words, it's a number we found through legitimate means, but it's not a solution to the original equation.

It's like if you put in an honest day's work and then got paid in Trident gum. You got something for your efforts, but it's not what you wanted, and it won't help you pay for anything. Although at least it does taste exactly like green apple and golden pineapple. Well, that's something.

Our final answer for this problem is the single solution x = 5.