If we had a time-traveling telephone booth, we could chat with
7th-century Indian mathematician Brahmagupta,
the 9th century Persian codifier of algebra al-Khwārizmī,
and the sixteenth-century Italian astronomer and physicist Galileo.
Why? Well, they have a lot in common.
In fact, in lots of times and places -- from ancient Babylon to Egypt and China --
we'd run into people talking about the same mathematical thing:
solving quadratic equations -- a special group of polynomials with x^2 as the largest power
of x that appears.
Quadratic equations grew from applications as diverse as humanity itself.
We use them to describe how much taxable grain can be grown in a field, the forces at work
on a falling object,
or even to experience the intellectual joy of solving a mathematical puzzle.
In fact, we still use quadratic equations today to model important scenarios in our lives.
G’Day, I’m James Tanton, and this is Study Hall Algebra, presented by Arizona State University
and Crash Course.
We’ve actually met quadratic expressions a few times already,
but we haven't been formally introduced.
So let’s fix that.
A quadratic equation is any polynomial that can be put into the form ax^2 + bx + c = 0,
where a, b, and c are all constants.
For example, x^2 + x = 6 is a quadratic equation because it can be rewritten in the form x^2+x-6=0
with a=1, b=1, and c = -6.
Even something like 3x^2 = 42 is a quadratic equation.
Zero is a constant, so we still have a quadratic equation if "b" and/or "c" are 0.
But if "a" were zero, there’d be no x-squared term.
And this is really the key piece.
Without an x^2, we’d be back to a linear equation like we solved way back in episode 1!
To visualize a quadratic equation with a graph, having an x^2 term is essential.
On a graph -- which we’ll get into next time -- quadratic equations are a particular
kind of U-shaped curve called a parabola,
and we need that x^2 term to make a parabola.
So today, as we're getting familiar with quadratic equations, we'll assume that “a” is never zero.
Now, last episode we talked about how polynomials in general can be used to turn complicated
mathematical models
into something we can work with more easily.
And, as we just saw on our whirlwind tour of history,
quadratic equations keep showing up in interesting real-life problems,
including problems with more than one solution.
Quadratic equations are especially great because they're simple enough that we can work with
them by hand,
yet complicated enough to help us practice the thinking we need to work with more difficult
polynomial equations later.
A less formal, but more historical definition is that a quadratic equation is one that can
be solved by drawing a square
– hence the name, which comes from the Latin word quadratum or quadratus, which means square.
Last episode we saw that the box or grid method is fantastically useful for multiplying and
dividing polynomials,
especially when we’re factoring, or figuring out what smaller polynomials multiply together
to produce our original polynomial.
It turns out quadratics are a special case where drawing a symmetrical quadrilateral is key.
Let’s start with x^2 – 36 = 0. We have a = 1, c = –36, and b … well it equals zero.
Remember from our definition, as long as "a" isn't 0, it's still a quadratic equation!
We can rewrite it as x^2 = 36, which will help us draw our square.
If we’re thinking of it in terms of geometry, we know that a square with an area of 36 square units
must have a side length of 6.
Then, we can build on that intuition to find what other values are algebraically true.
We know -6 squared also gives us 36.
So this quadratic equation has two roots, or solutions that will make the sentence true:
x = -6 and x = 6.
The roots are sometimes called zeroes because they’re the values that make our original
quadratic expression 0.
Many of the quadratics you'll find in algebra class have two real roots (as in not imaginary numbers).
And this is the maximum, because a number has at most 2 square roots.
But there are some special cases too, like x^2=0.
This is still a quadratic equation, but there is only one number whose square is zero.
It's 0! So here there’s just one real root: x = 0.
Some quadratics even have no real solutions.
Remember, equations like x^2 = –7 could only be solved using the imaginary number i.
So every quadratic equation will have either zero, one, or two real solutions.
But finding those solutions and the square hidden in our quadratic equation can often be tricky.
Take x^2 – 3x + 1 = 0. If we try to factor by guessing and checking,
it becomes clear pretty quickly that we can’t solve this with integers.
There are no integers we can add to get –3 and multiply to produce 1.
So if we typed this quadratic into a calculator to solve it, without quite knowing what we're doing,
we’d end up with weird infinite decimals, which means there are square roots lurking
in there somewhere.
We’ve been cornered into having to really solve this quadratic to get precise answers!
So what do we do?
One strategy is to use careful detective work and rely on the symmetry of the square.
Let’s look at each term (or piece of evidence) in the quadratic equation one at a time.
The x^2 is most important, so let’s start there.
Sure, it could come from something like x/2 * 2x, but let’s stick with symmetry and
avoid unnecessary fractions.
So we can start with x times x.
Now we need to bring in the -3x for questioning.
We have to split it into two symmetrical parts, one on each side of the square.
We could have a -3/2 on each side, but although that would be mathematically correct, fractions
just feel complicated.
My human brain prefers integers.
To keep our mathematical sleuthing simpler, remember that an equivalent equation will
have the same solutions.
Let's multiply our original quadratic equation by something to make a nicer equivalent equation.
If we multiply the original equation by 2, we’d get 2x^2 - 6x + 2 = 0.
Then we'd have to deal with splitting up 2x^2, which seems equally complicated and full of
square roots of two.
So I don't want to. Let's try something else.
Instead, let’s multiply the original equation by 4, which gives us 4x^2 – 12x + 4 = 0.
Aha! Now we have an equivalent equation with a nice middle term and a nice first term.
So if we draw our square and review each piece of evidence separately, the problem seems
more manageable.
4x^2 can be split up symmetrically into 2x times 2x.
And we can also easily split the -12x piece symmetrically into -6x + -6x.
Those could each come from 2x times -3.
Now we just have to do one more thing to complete the square. -3 times -3 is 9.
Wait, 9? We weren’t going for 9!
This is the tricky bit, because it’s rare that we’ll end up with exactly what we wanted
in that final box.
But we can work with 9. We just need to modify our equivalent equation a little bit more.
At the moment, the equation we’ve got is 4x^2 – 12x + 4 = 0.
But our square is telling us that the best expression to have would be 4x^2 –12x +
/9/.
We can make that happen by adding 5 – we just have to be careful to add it on both
sides of the equals sign.
The left side is precisely our square, 2x-3 times 2x-3, and our equivalent equation tells
us the area is 5.
So (2x-3) squared is 5, meaning 2x-3 must equal positive or negative sqrt(5).
In one final investigative flourish, we can use these to solve for x.
Either 2x equals 3 + √5 or 2x equals 3 – √5.
So our two solutions here, our two real roots, become (3 + √5)/2 and (3 – √5)/2.
This sort of multi-step detective work isn't the simplest,
but it can help us solve every single quadratic equation in the world,
whether it has nice neat integer solutions or messy complex number solutions.
Now you might be wondering why we haven’t yet talked about the famous quadratic formula.
Surely we should have it framed on the wall, or have composed a glorious ode to memorize it.
It certainly doesn’t hurt to know it, and it can be useful as a shortcut.
But believe it or not, we don’t actually need it. We’ve been using the original quadratic
formula all along – the quadratus itself.
We actually have all the skills we need to derive the formula ourselves by completing
the square for ax^2 + bx + c = 0.
Try it!
In fact, completing the square works for all types of quadratic equations.
The classic example of “quadratics in the real world” is projectile motion.
But these examples end up being pretty theoretical, and you really don’t need to sit down with
a calculator before tossing a tennis ball for your dog.
Something a little higher-stakes than a game of fetch is calculating braking distances
for cars,
which some auto manufacturers and road authorities approximate with a quadratic equation.
Let’s take a closer look at this through the Math Goggles.
Let’s say that we're driving down a dark country road, and the furthest our headlights
can illuminate clearly is 120 feet.
We want to find the max speed we can be driving and still stop in that braking distance.
The physics of a braking car are complex, but the British Highway Code has simplified
it to the model d = (s^2)/20 + s.
"s" is the speed you are going in miles per hour and "d" is your stopping distance in
feet… approximately.
The “+ s” comes from the bit you coast before your feet actually react to hit the
brake,
and physics shows that the actual stopping distance after that is a constant times the
speed squared.
Going back to our night drive and our question about maximum speed,
we can insert 120 feet as "d" and get the quadratic equation (s^2)/20 + s = 120.
To start, let's multiply by 20 to avoid that pesky fraction and get the equivalent equation
s^2 + 20s = 2400.
Then, let’s set up our square. s times s is s^2.
We’ll split 20s into 10s and 10s. And 10 times 10 gives us 100,
which means we have to make another equivalent equation by adding 100 to both sides.
After all that, we have (s + 10)^2 = 2500.
Taking the square root, (s + 10) is either 50 or –50. So s is either 40 or –60.
We have to think about what our solutions mean.
The positive answer makes physical sense, but the negative answer would mean we're driving
backwards in the dark at break-neck speed.
...Don’t do that.
So if we’re going 40 miles per hour, it takes all 120ft to brake.
So on an unlit country road, we’d better play it safe and go a little less than 40
if we want to be able to stop in time.
Solving quadratic equations can feel intimidating.
But you already have all the tools you need without memorizing anything – your built-in
human inventiveness, logic, and eye for symmetry.
If you’re stuck, the first, the best, the greatest tip for working with a quadratic
equation is to draw a square and complete it (literally!).
Next time we’ll look at what it really means to graph an equation and how we can use visual
thinking to help us set up and solve algebraic equations.
Hey, that sounds like what we did when we looked at word problems.
It’s almost like all of mathematics is connected by human experience and ingenuity...
Thanks for watching Study Hall Algebra, which is produced by Arizona State University and
the Crash Course team at Complexly.
If you liked this video and want to keep learning with us here in Study Hall, be sure to subscribe.
You can learn more about ASU and the videos produced by Crash Course
in the links in the description.
Drive safely, and see you next time!