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Squares and Square Roots

Squares and Square Roots

Completing the Square

Now we know how to solve a quadratic equation that can be rearranged so we have the square of a first-degree 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. But what are we gonna 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: x2 + 2x – 4 = 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 a polynomial. Hmm...let's try factoring the left side as (x + 1)2 and see what we get.

(x + 1)2 = x2 + 2x + 1

Nope, that won't work. The equation we're supposed to solve has "minus 4" instead of "plus 1." Strike two. What do we do, other than wait for a better pitch?

First, add 4 to each side to get that number out of the way.

x2 + 2x – 4 = 0
x2 + 2x = 4

If we had "plus 1" on the left-hand side, we could rewrite that side of the equation as the square of a first-degree 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:

x2 + 2x + 1 = 4 + 1

Now we can factor the left-hand side of the equation and add the numbers on the right-hand side.

(x + 1)2 = 5

We know how to solve this bad boy. Rock some square rootin':

Then, we subtract 1 from both sides:

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 x2 and x terms of something that's supposed to be a square.

Sample Problem

In order for x2 + 8x + c to be the square of a binomial with degree 1, what must c equal?

If x2 + 8x + c = (x + ?)2, then 2(?) = 8, so ? must equal 4. If that's true, we can find c by multiplying it out:

(x + 4)2 = x2 + 8x + 16

Sweet. 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 x2 is 1, to find c we divide the coefficient of x by 2, and then square it. Oh. That wouldn't make for a great power tie, but whatever.

Sample Problem

If x2 + 6x + c is a perfect square, find c.

We divide 6 (the coefficient on x) by 2:

6 ÷ 2 = 3

Then square that answer:

32 = 9

That means c = 9, so the expression x2 + 6x + 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 x2 + 3x + 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 .

Bam, that means .

This can become trickier if the coefficient on x2 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 9x2 + 66x + c is a perfect square, what's the value of c?

We know that (px + q)2 = 9x2 + 66x + c, but we don't know what p or q are yet.

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

In order for (px)(px) to be 9x2, the coefficient on each individual x must be 3, so p = 3.

Now we know that:

(3x + q)2 = 9x2 + 66x + c

We still need to find q. It's not in our box of alphabet blocks, which is where we saw it last. Using the distributive property, we know that the middle term is 66x. And since that middle term is the sum of (3x)(q) and (3x)(q), we can track it down:

(3x)(q) + (3x)(q) = 66x
2(3x)(q) = 66x

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

(3x + 11)2 = 9x2 + 66x + 121

The coefficients on the x2 and x terms match what we were given originally, and now we know that c = 121. You can't hide from us, c.

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: x2 + 14x + 10 = -35.

First, we move all the constants to the right-hand side of the equation. Otherwise, they'll constantly be in our way and we'll probably trip over them.

x2 + 14x = -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 x2 + 14x + c is a square. The coefficient on the x term is 14, so let's zoom in on him for a sec. Since 14 ÷ 2 = 7 and 72 = 49, we need to choose c = 49. This turns our equation into:

x2 + 14x + 49 = -45 + 49

Factor the left-hand side and simplify the right-hand side:

(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 x2 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'll result in our coefficient on the x2 being 1, and continue as normal.

Sample Problem

Solve: 2x2 + 10x = 4.

The x2 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:

x2 + 5x = 2

Now the coefficient on x2 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 it to find the number we need to add to both sides. Warning: it'll 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. In case you forgot which side is which: righty-tighty, lefty-loosey. Wait, that only helps if you're trying to open a bottle of Snapple. Let's try that again:

Ah, now we have a first-degree 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 the square root:

Simplify the square root:

Then, solve the two equations:

And 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 1000-piece puzzle that's been taking up space in your basement for the past three months. We can always divide  both sides of a quadratic equation to make the coefficient of x2 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 x2 + 6x = 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 picking 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:

x2 + 6x + 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.

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