# At a Glance - Taking Square Roots

The easiest possible quadratic equation to solve would be something like x2 = 9, so we won't be dealing with anything more complicated than that.

Psych! (Sorry.)

If the square of x is 9, then x can be the positive or negative square root of 9:

x2 = 9

This simplifies to:

x = ±3

All we need to do is remember to keep both the positive and negative square roots. Sometimes after taking the square root, though, we need to do a little more work. Put your back into it.

### Sample Problem

Solve: (x + 3)2 = 9.

If the square of x + 3 is 9, then x + 3 can be either the positive or negative square root of 9. In other words, we're taking the square root of both sides to knock out that exponent.

(x + 3)2 = 9
x + 3 = ±3

Now we need to solve two equations. Quickly, before the amount of work we need to do doubles again:

x + 3 = 3 and x + 3 = -3

For the first equation, if x + 3 = 3, then x = 0.

We're always paranoid that we might have done something wrong, so we'll check to make sure that this works. When we plug in x = 0 to the left-hand side of the original equation, we find:

(0 + 3)2 = 9

Yep, that's true. So x = 0 is a solution to the equation. Halfway home.

Now for the second equation. If x + 3 = -3, then x = -6.

Uh-oh. Our paranoia is setting in again, so it's time to check to see that this works. When we plug in x = -6 to the left-hand side of the equation, we get:

(-6 + 3)2 = 9
(-3)2 = 9

Hey, that's also true. Therefore, x = -6 is also a solution to the equation.

The final answer: both x = 0 and x = -6 are solutions to the equation. There's room in this town for the both of them.

The examples we've done so far have had a squared variable on one side and a number on the other. If we have an expression that's a square multiplied or divided by something, we need to "undo" that multiplication or division before taking the square root. Unfortunately, we can't just hit "Control Z."

### Sample Problem

Solve: 2x2 = 72.

First we divide both sides by 2 to find an equation that has a square on one side:

x2 = 36

Then we can take the square root of 36:

x = ±6

Sometimes we might see an equation has the square of a first-degree polynomial on one side, but the square is in disguise and needs to be factored. Watch for trench coats and low-brimmed hats.

The last few sample problems were all well and good, but solutions to quadratic equations don't always result in beautiful numbers. For example, take the ugly side of quadratic equations. Try not to look directly at them, though, since we don't want you to go blind.

Of course, we could put together all the things that can happen to make something horribly complicated, and trust us, we will. Cue maniacal laugh.

### Sample Problem

Solve 18x2 + 12x + 2 = 7.

First, we can factor the left-hand side:

2(9x2 + 6x + 1) = 7
2(3x + 1)2 = 7

We want a square on one side of the equation, so divide both sides by 2:

Now that we have a square on one side, we can take the square root of both sides:

We solve the two equations:

Hit both of 'em individually to get our two solutions:

These solutions can be rewritten, or simplified:

These look like a nightmare you might have after consuming too much dairy right before bed, but they're absolutely correct. In fact, check the first solution. Plugging it into the original equation would be the ideal thing to do, but this isn't an ideal world, and that arithmetic would be too messy. Instead, since we're confident we rearranged the equation correctly at the beginning, plug the solution into (3x + 1)2 and make sure we get . Here goes:

Made it, Ma! Top of the world!

The general plan of attack for solving this sort of equation is to attack quickly and aggressively from the left flank. No, wait, that's the general plan for attacking a right-handed enemy in a mountainous region. Never mind us.

The real plan of attack is to manipulate the equation so there's something squared on one side and a number on the other side, even if it gives you something truly awful to look at. Take the appropriate square root, and then solve each new equation you find. Remember, you can always check that your answers are right by plugging them back into the original equation and making sure they work. If they don't, try replacing the batteries before giving up. They take Triple-As.

Of course, this won't work for a quadratic equation that has no real number solutions, like the equation (x + 1)2 = -3.

This can't possibly have real number solutions, since there's no way to square a real number (x + 1) and end up with -3. We've tried. A lot. Can't be done.

#### Example 1

 Solve: x2 + 2x + 1 = 16.

#### Example 2

 Solve: x2 = 5.

#### Example 3

 Solve (x – 6)2 = 15.

#### Example 4

 Solve -2x2 = 16.

#### Exercise 1

Solve or determine that the equation has no real number solutions:

y2 = 16

#### Exercise 2

Solve or determine that the equation has no real number solutions:

7x2 = 63

#### Exercise 3

Solve or determine that the equation has no real number solutions:

(x – 2)2 + 16 = 0

#### Exercise 4

Solve or determine that the equation has no real number solutions:

4(x + 1)2 = 36

#### Exercise 5

Solve or determine that the equation has no real number solutions:

#### Exercise 6

Solve or determine that the equation has no real number solutions:

3x2 – 7 = 11

#### Exercise 7

Solve or determine that the equation has no real number solutions:

(2x + 5)2 = 36

#### Exercise 8

Solve or determine that the equation has no real number solutions:

3x2 = 4

#### Exercise 9

Solve or determine that the equation has no real number solutions:

x2 + 6x + 9 = 100

#### Exercise 10

Solve or determine that the equation has no real number solutions:

4x2 + 12x + 9 = 20