# Squares and Square Roots

### Topics

Now we know how to solve a quadratic equation that can be rearranged so we have the square of a degree 1 expression on one side and a number on the other side. We also know how to solve quadratic equations when things factor nicely on one side and there's a zero on the other side. However, whatever shall we do we do with equations that don't fit either form? Other than spread our arms upward toward the heavens in utter despair and weep uncontrollably?

The answer, of course, is lots of things.

### Sample Problem

Solve: *x*^{2} + 2*x* – 3 = 0.

We can't factor this nicely, which is strike one. It almost looks like we could rewrite the left-hand side of the equation as the square of degree 1 polynomial *x*. However,

(*x* + 1)^{2} = *x*^{2} + 2*x* + 1

and the equation we're supposed to solve has "minus 3" instead of "plus 1." Strike two. What do we do, other than wait for a better pitch?

First, add 3 to each side to get that number out of the way. Now we have

*x*^{2} + 2*x* = 3.

If we had "plus 1" on the left-hand side, we could rewrite that side of the equation as the square of a degree 1 polynomial. We can't add 1 to only one side of the equation, but we *can* add 1 to both sides of the equation, like this:

*x*^{2} + 2*x* + 1 = 3 < + 1.

Now we can rewrite the left-hand side of the equation, and add the numbers on the right-hand side of the equation to find

(*x* + 1)^{2} = 4.

We know how to solve this bad boy. Taking a square root yields

*x* + 1 = ± 2.

Then, we solve the equations

*x* + 1 = 2 and *x* + 1 = -2

to find, respectively, the solutions

*x* = 1 and *x* = -3.

We can check these answers by plugging them into the original equation. First, check *x* = 1:

(1)^{2} + 2(1) – 3 = 3 – 3 = 0,

so *x* = 1 works. Ball one. Sorry, we felt like we needed to stick with the baseball analogy. Next, check *x* = -3:

(-3)^{2} + 2(-3) – 3 = 9 – 6 – 3 = 0,

so *x* = -3 works also. Ball two. Take your base. We know it takes four balls for a walk, but we can't keep this going any longer, so smile and nod.

What we did is called **completing the square**. First, we added to both sides of the equation a number that allowed us to look at one side of the equation as the square of a polynomial with degree 1. From there, we knew how to solve the equation, so we went ahead and did so.

Thanks, completing the square. You're a lifesaver.

The trick is to figure out the missing constant term when we're given the *x*^{2} and *x* terms of something that's supposed to be a square.

### Sample Problem

In order for *x*^{2} + 8*x* + *c* to be the square of a binomial with degree 1, what must *c* equal?

If *x*^{2} + 8*x* + *c* = (*x* + ?)^{2},

then 2(?) = 8, so ? must equal 4. Since

(*x* + 4)^{2} = *x*^{2} + 8*x* + 16,

we know that *c* = 16.

We sense a pattern here. Hmm. It's not paisley, it's not houndstooth...what is it? When the coefficient on *x*^{2} is 1, to find *c* we divide the coefficient of *x* by two, and then square. Oh. That wouldn't make for a great power tie, but whatever.

### Sample Problem

If *x*^{2} + 6*x* + *c* is a square, find *c*.

We divide 6 (the coefficient on *x*) by 2 to find 3, then square to find 9. The expression

*x*^{2} + 6*x* + 9

factors as

(*x* + 3)^{2}.

If the coefficient on *x* is odd, we follow the same pattern.

### Sample Problem

What do we need to square to find *x*^{2} + 3*x* + c, and what is *c*? And no, "The third letter of the alphabet" is not an acceptable answer.

Divide 3 by 2. If we square

,

we find

.

Therefore, we square

and .

This can become trickier if the coefficient on *x*^{2} isn't 1. Since a little bird told us that "tricky" is your middle name, we'll go ahead and show you how that would work.

### Sample Problem

Assuming the expression below is a square, find *c*: 9*x*^{2} + 66*x* + *c*.

We know that (?*x* + ?)^{2} = 9*x*^{2} + 66*x* + *c*.

What are the missing numbers? We're worried. Has anyone tried posting their pictures on the side of a milk carton?

In order for (?*x*)(?*x*) to be 9*x*^{2}, the coefficient on each individual *x* must be 3: ? = 3.

Now we know that

(3*x* + ?)^{2} = 9*x*^{2} + 66*x* + *c*.

We still need to find ?. It's not in our box of punctuation marks, which is where we saw it last. Using FOIL, we know that the sum of the outside and inside products, which are the same, is 66*x*:

2(3*x*)(?) = 66*x*.

The missing ? must be 11. We'll check to see if we're right:

(3*x* + 11)^{2} = 9*x*^{2} + 66*x* + 121.

The coefficients on the *x*^{2} and *x* terms match what we were given originally, and now we know that *c* = 121.

When we have an expression

*ax*^{2} + *bx* + *c*,

that's the square of an expression

(?*x* + ?).

We need to eliminate those question marks, so we can stop feeling like the expression is giving us the third degree. We can figure out ? by taking the square root of the coefficient *a*. Then we divide *b* by 2*a* to figure out ?. Now those question marks can get out of our hair and go somewhere they'll be more useful, like a police interrogation room.

Once we know what we're squaring to find the expression

*ax*^{2} + *bx* + *c*,

we can figure out *c*. Once we know *c*, the sky's the limit. Well, *c* is the limit. Once we find that, we're done.

Now that we're experienced at figuring out the missing constant term in a squared expression, it's time to use this new skill to solve some more equations. Once we've done that, it's on to solving world hunger. One thing at a time.

### Sample Problem

Solve: *x*^{2} + 14*x* + 10 = -35.

First, we move all the constants to the right-hand side of the equation. Otherwise, they will constantly be in our way.

*x*^{2} + 14*x* = -45

Then we complete the square. We want to add a constant *c* to both sides of the equation, and we want to choose *c* so that *x*^{2} + 14*x* + *c* is a square. Therefore, we need to choose *c* = 49. This yields

*x*^{2} + 14*x* + 49 = -45 + 49 ,

and factoring the left-hand side and simplifying the right-hand side gives us

(*x* + 7)^{2} = 4.

This gives us two equations to solve:

*x* + 7 = 2 and *x* + 7 = -2.

The solutions are, respectively,

*x* = -5 and *x* = -9.

While we can complete the square even if the coefficient on the *x*^{2} isn't 1, that's more trouble than it's worth when solving equations. Instead, we divide both sides of the equation by the number that will result in our coefficient on the *x*^{2} being 1, and continue as normal.

### Sample Problem

Solve: 2*x*^{2} + 10*x* = 4.

The *x*^{2} term has a coefficient of 2, which is 1 too many. Divide both sides of the equation by 2, and force it to be the number we want it to be:

*x*^{2} + 5*x* = 2.

Now the coefficient on *x*^{2} is 1, so it's easier to complete the square. The coefficient on the *x* term is 5, so divide 5 by 2 and then square to find the number we need to add to both sides. Warning: it will probably be a fraction.

Don't say we didn't warn you. Now factor the left-hand side of the equation and add the numbers on the right-hand side of the equation. In case you forgot which side is which: righty-tighty, lefty-loosey. Wait, that will only help if you're trying to open a bottle of Snapple. Let's try that again:

Ah, now we have a degree 1 polynomial squared on the left, and a number on the right. That's the way (uh-huh uh-huh) we like it. We know what to do from here. Take a square root:

.

Simplify the square root:

.

Then, solve the two equations

to find the solutions

.

Completing the square will let us solve any quadratic equation that has real numbers as solutions. Plus, it feels so good to "complete" something unlike that 1,000-piece puzzle that's been taking up space in your basement for the past three months. We can always divide through both sides of a quadratic equation to make the coefficient of *x*^{2} be 1, and we can always divide the coefficient of *x* by 2 and square to find the right constant to add to both sides of the equation.

The only step where things could go wrong is the one where we take square roots. If we try to take the square root of a negative number, we'll run into problems. We hate running into *anything*. Hurts like the dickens.

The equation (*x* + 1)^{2} = -4, for example, has no solutions. As long as we're taking square roots of non-negative numbers, however, completing the square will always work. It's the only thing as certain as death and taxes.

### Sample Problem

A student was asked to solve the equation *x*^{2} + 6*x* = 16.

She was a little distracted by a blue jay outside the classroom window, however, and wrote down the following work. What did she do wrong, aside from pick an inappropriate time to birdwatch?

Basically, our amateur birdwatcher forgot to add 9 to both sides of the equation. The first line should have been

*x*^{2} + 6*x* + 9 = 16 + 9.

Maybe next time she'll keep her eye on her paper, where it belongs. She can doodle a pigeon in the margin if it helps.