A quadratic equation is an equation of the form:
ax2 + bx + c = 0
In other words, it's an equation that sets a quadratic polynomial equal to zero. Got that? On the left side of the equal sign, a quadratic polynomial, and on the right side...well, there ain't no thang.
We've solved some quadratic equations already by factoring to find the roots of quadratic polynomials. Now we'll dive more deeply and learn how to solve quadratic equations that can't be easily factored. Don't go too fast or you'll get the bends.
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.
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.
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."
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
Let's kick things off with a...
What, you were expecting a party?
Solve the quadratic equation ax2 + bx + c = 0 by completing the square.
Here we know a, b, and c are numbers, but we don't know what any of them are. We do know that a can't be 0, or we wouldn't have a quadratic equation. We have a sneaking suspicion that b is 17, but that's only based on a dream we had last night, so we should probably do the math to be on the safe side.
First, we divide both sides by a, since we don't want a coefficient on x2 gumming up the works.
Then we subtract from both sides to get it out of the way.
Next we take (the coefficient on x), divide by 2, and square to find . It's not as nice-looking as what we've had in the past, but we'll go with it. This is the number we add to both sides:
The left-hand side of the equation can now be written as a square:
Since , we can add the numbers on the right-hand side of the equation:
It's a good feeling to look at such an ugly conglomeration of numbers, variables, fractions, and parentheses and know that you can make some sense out of it, eh? Not that you want to print it out and hang it on your bedroom wall, but still. We think it's nice.
Now, writing the left-hand side of the equation as a square and the right-hand side of the equation in its new form, we need to solve the equation:
Since we have the square of a first-degree polynomial on the left and an admittedly messy number on the right, we know where to go from here. We summon our radical powers and take the square root of both sides.
We can simplify the right-hand side:
Now, we need to solve these two equations:
Here's our first solution:
Ready for the second solution? Comin' atcha:
These solutions are usually written together as:
We call this gnarly equation the quadratic formula.
You may have already heard of this one. It's famous. It used to have its own talk show, and it was voted one of People's Sexiest Formulas Alive in 300 BCE.
Okay, maybe it hasn't achieved that kind of fame, but it is well-known. The quadratic formula is another way we can find solutions to quadratic equations. When given a quadratic equation, we figure out which numbers correspond to a, b, and c, then plug them into the quadratic formula to find our answers. It's like one of those factory machines where you throw in various ingredients, and then out comes a gloriously wonderful Oreo, or Twinkie, or wax lips. Yes, we went with "wax lips."
As complicated as it is to arrive at and understand the quadratic formula, it's one of those things that's best to memorize and use. It's unlikely that you'll be asked in school to explain where the quadratic formula comes from, but it's reassuring to know that, like babies, it does come from somewhere.
Try to make it feel at home, won't you?
Solve: x2 + 5x – 7 = 0.
The coefficient on x2 is 1, the coefficient on x is 5, and the constant term is -7, so we have:
All we do is plug these numbers into the quadratic formula, , and we'll have our solution in no time:
Since 53 is a prime number, we can't simplify any more. The two solutions to the equation are:
Be Careful: In order to use the quadratic formula, you need an equation of the form ax2 + bx + c = 0. Because we must have zero on one side of the equation, you can't go quadratic equation-happy and start plugging everything under the sun into it. By the way, if you ever go overseas, make sure you pack your European quadratic plug adapter.
A student, the same hopeless daydreamer who aspires to one day be a member of the National Audubon Society, was asked to use the quadratic formula to solve the equation x2 + 3x – 6 = 7.
She wrote down the following work. What did she do wrong?
The original equation didn't have zero on one side. D'oh! The student should have first subtracted 7 from each side, then solved the equation x2 + 3x – 13 = 0.
This is starting to become a problem. Her teacher might want to consider drawing the blinds.
A different student, one who's not interested in birds so much as he is captivated by the incoming texts on the cell phone he's discreetly concealing in one of his sleeves, was asked to use the quadratic formula to solve the equation 4x2 + 15x – 9 = 0.
He wrote down the following work. What did he do wrong?
In the original equation, c was -9, not +9. He might have noticed that if he hadn't been so busy lol'ing and jk'ing.
The student's first two lines should have been:
Sometimes a quadratic equation has no solutions. The quadratic equation says . But since we can't take square roots of negative numbers when we're looking for real number solutions, this only works if b2 – 4ac is positive (or zero). The expression b2 – 4ac is called the discriminant of the equation ax2 + bx + c = 0.
The quantity b2 – 4ac discriminates between those equations that have real number solutions and those that don't. That's a whole lot of power for one tiny little formula.