# The quadratic formula: a tutorial

So you need to solve $2x_{2}−3x=2$. You remember the word ‘quadratic’ and something like $4ac−b_{2} $ … or was it $+b_{2}$? 😨

Stop trying to memorize the quadratic formula! In this interactive tutorial, you’ll *rediscover* the quadratic formula with me, and you’ll never have to memorize it again! 💪

First let’s remind ourselves of the problem that the quadratic formula solves. A **quadratic equation** is one that looks like this, for some particular constants $A$, $B$ and $C$:

For example, $2x_{2}−3x=2$ is a quadratic equation. What are its values of $A$, $B$ and $C$?

Careful with the minus signs. Rearrange $2x_{2}−3x=2$ into the form $Ax_{2}+Bx+C=0$. You can do this by subtracting $2$ from both sides. You end up with $2x_{2}+(−3)x+(−2)=0$.

Good, you paid attention to the minus sign! 😁

‘Solving’ a quadratic equation means finding the values of $x$ that make the equation equal on both sides. These special values of $x$ are called the **roots** 🥕 of the equation. Consider again our quadratic equation, $2x_{2}−3x−2=0$. Which of these is **not **a root?

Actually, $x=−21 $ **is** a root, as you can see by plugging it into the equation:

Actually, $x=2$ **is** a root, as you can see by plugging it into the equation:

By plugging a value of $x$ into a quadratic equation, you can quickly *check* whether that value is a root. Above, plugging in $x=2$ or $x=−21 $ gives the valid equation $0=0$, so they are both roots. But plugging in $x=−2$ gives the invalid equation $12=0$, so it is not a root.

Unfortunately, trying random values of $x$ is not an efficient way to find the roots!

**Graphing the equation 📈**

Instead of plugging in *random *guesses, we’d rather plug in *sensible* guesses. The **graph **is a helpful tool for finding sensible guesses. (Later, we’ll find that the graph is helpful for deriving the quadratic formula, too!)

Let’s solve another quadratic equation: $−1x_{2}−1x+2=0$. As we go, we’ll plot our guesses on a graph, like this:

Here, we’re plotting $y=−1x_{2}−1x+2$. Our goal is to find an $x$ where $y=0$, as this would solve our equation. We’ve made a couple of guesses already at the $✘$ marks. What guesses have we made?

Not quite: those are the $y$-values. We’ve made two guesses, $x=0$ and $x=2$, for which we calculated the $y$-values $y=2$ and $y=−4$. For $x=0$, we calculated $y=2$:

In the same way, for the guess $x=2$, we calculated $y=−4$:

Right, we calculated two points: $(x=0,y=2)$ and $(x=2,y=−4)$.

These two guesses were unsuccessful: they did not solve the equation $−1x_{2}−1x+2=0$. What would a successful guess — that is, a **root **— look like on our graph?

It’s the other way around. The $y$-axis line is the set of points where $x=0$, so a $✘$ on the $y$-axis would just be the guess $x=0$.

We’re trying to solve $−1x_{2}−1x+2=0$. We’re plotting $y=−1x_{2}−1x+2$, so finding a point where $y=0$ would solve our equation.

The $x$-axis line is the set of points where $y=0$. So all $✘$ on the $x$-axis are roots of our equation.

Using the graph, our incorrect guesses can help us choose a better next guess. To see why, plot $y=−1x_{2}−1x+2$ for the rest of the range $−5≤x≤5$. Which of the following curves does it look like? (You might want to grab paper and pencil.)

It’s not curve $B$. Try for example $x=−3$, where you should get $y=−4$. This is far from the point $(x=−3,y=4)$ on curve $B$. If you plot all points correctly, you’ll find they follow curve $C$ above.

It’s not curve $A$. Try for example $x=5$, where you should get $y=−28$. This is far from the point $(x=5,y=−3)$ on curve $A$. If you plot all points correctly, you’ll find they follow curve $C$ above.

After plotting curve $C$, its roots should be clear. What are they?

No, that’s the intersection of $C$ with the $y$-axis. We want the intersection of $C$ with the $x$-axis. These intersections are at $x=−2$ and $x=1$.

That’s one root, but there’s another at $(x=−2,y=0)$.

Notice that curve $C$ follows an upside-down ‘bowl’ shape 🥣. If you plot some more quadratics, you’ll find that they all follow a similar ‘bowl’ or ‘U’ shape. It can be flipped upside down, or shifted up or to the side, or squished horizontally or vertically. But it’s always a ‘bowl’ shape, called a **parabola.**

Which of the following curves could **not **possibly be a quadratic equation?

No, that could actually be a quadratic! Only $D$ is definitely not quadratic.

Well done. $D$ could not be a quadratic, because it has an ‘S’ shape.

$A$ and $C$ are both **parabolas **(‘bowl’ shapes, where bowl $A$ happens to be upside-down). $B$ could also be part of a very stretched parabola. But the curve $D$ has an ‘S’ shape, so it could not be quadratic.

Knowing that a quadratic always forms a ‘U’ shape helps us make better guesses. In the following graph of some quadratic, we’ve already made three guesses:

Considering the graph must be a parabola, what is a good next guess?

No, this could not work. Only a ‘U’ shape could fit through these three points, and the only $x$-value that looks like it could work is $x=2$:

Right! Only an upwards ‘U’ shape could fit through these three points, like so:

So graphs help us make better guesses, but we’re still just guessing. This is not efficient either! Instead, the **quadratic formula** will give us an efficient way to find *all *the roots of *any* quadratic equation.

**Revisiting multiplication**

Here’s a simpler equation with two unknowns, $P$ and $Q$:

What best describes **all **possible solutions to this equation?

This does satisfy $P×Q=0$, but there are more solutions! For example, $P=5$ and $Q=0$. You can visualize all the possible solutions by writing them out in a big table:

The result is zero all down the $P=0$ column and all along the $Q=0$ row. So as long as one of the values is zero, the other can be anything.

Right! As long as one is zero, the other can be anything.

Now here’s a similar equation, with two different unknowns $L$ and $R$:

As you’ll discover shortly, this is a quadratic equation! But what are its solutions?

No, it’s actually just like the last equation! We can let $P=x−L$ and $Q=x−R$. If $P=0$, then $x−L=0$. If $Q=0$, $x−R=0$.

Right, it’s just like the last equation!

You can simplify those solutions a bit. With a bit of rearranging, what do you get?

Be careful with the minus signs. Take $x−R=0$ and add $R$ to both sides. You should get $x=R$.

** **

**Multiplying out**

The equation $(x−L)(x−R)=0$ is actually a quadratic! You might remember how to ‘multiply out’ brackets, like this:

Using this method, multiply out $(x−L)(x−R)$.

Almost. Be careful with the minus signs. You should set

Multiplying out, you should get

Right. The negatives cancel when multiplying, and we can factor a $−x$ out of the two middle terms.

So if I gave you this equation:

What are its roots? That is, what values of $x$ solve this equation?

No, it’s simpler than that! Earlier you found the solutions to $(x−L)(x−R)=0$ are $x=L$ or $x=R$. And you found that $(x−L)(x−R)$ can be rearranged as $x_{2}−(L+R)x+LR$. Rearranging an equation doesn’t change its solutions, so the solutions are still the same: $x=L$ or $x=R$.

Still don’t believe me? Plug in $x=L$ and verify:

You can check the same with $x=R$.

Yes! We just rearranged the previous equation, so the solutions are still the same.

Hopefully, the above equation looks familiar …

Yes, it’s a quadratic equation! But what are its values of $A$, $B$ and $C$?

Almost. Just pay attention to the minus signs: $B=−(L+R)$.

This is almost a general quadratic equation, except that $A=1$. For now, we’re going to deal with quadratic equations of the more restricted form:

The lowercase $b$ and $c$ signal this restricted form where $A=1$. Later, we’ll see how to deal with different values of $A$.

If we graph the equation $y=x_{2}−(L+R)x+LR$, which of the following could it look like?

Right!

No, it can’t be curve $B$ because that has no solutions — but we found earlier that $(L+R)x+LR=0$ must have two solutions, $x=L$ and $x=R$.

It can’t be curve $A$ because inverted ‘U’ shapes only happen with $A<0$. To see why, think about extreme values, like $x=±1000$. Out here, $x_{2}$ becomes huge and the other terms become irrelevant.

Curve $C$ is the only possible curve for a quadratic with $A=1$, because it’s the only upwards ‘bowl’ curve that crosses the $x$-axis. The curve of $y=x_{2}−(L+R)x+LR$ must follow a parabola, crossing the $x$-axis at the $Left$ and $Right$ root:

**Finding the minimum 📉**

We’re almost there! Instead of finding $L$ and $R$ directly, we’re first going to find the value exactly *between *them. Now consider the following graph of $y=(x−1)(x−3)$:

By now, you know how to find $x$ values that make $y=0$. Above, those solutions are $x=L=1$ and $x=R=3$.

But now a different question: in the graph above, what value of $x$ above *minimizes* the value of $y$?

Maybe you had the right idea looking at the point $(x=2,y=−1)$, where the bottom of the ‘U’ shape is. However, I was looking for the $x$ coordinate, $x=2$.

No, that minimizes $x$, but certainly doesn’t minimize $y$. The $y$ value becomes *larger *as the $x$ value goes towards $−∞$.

We’ll call this minimum $x$ value $M$. We can find $M$ just by looking at the graph above. First we find the bottom point of the ‘bowl’, at $(x=2,y=−1)$. Then $M$ is its $x$ coordinate, so $M=2$:

This curve is beautifully symmetrical, isn’t it? It’s an exact mirror image along the $x=M$ curve, with the roots $L$ and $R$ on either side. This symmetry will be the key to unlocking the quadratic formula!

**Averages 🔔**

Because the curve is symmetrical, the minimum $M$ is exactly between $L$ and $R$. That is, $M$ is the *mean* (or average) of $L$ and $R$.

As a reminder, the mean $M$ of any two numbers $L$ and $R$ is defined as

For example, if $L=3$ and $R=7$, what is their mean $M$?

You’re one off. Try a calculator: $23+7 =5$.

$5$ is exactly in the middle of $3$ and $7$, as we can visualize on a number line:

That distance $d=2$ is the same on both sides. (That’s what it means for $M$ to be ‘in the middle’!)

What’s interesting is that we can define these the other way around. Say $M=4$ and $d=6$. Then what are $L$ and $R$? (Assume that $L≤R$.)

Not quite; you probably got $M$ and $d$ the wrong way around. Here’s the correct calculation:

In general, we can write:

Remember that our goal is to find $L$ and $R$, the roots of our quadratic equation, $x_{2}+bx+c=0$. But our strategy will be to first find $M$ and $d$, then plug those into the equations above to find the roots $L$ and $R$.

**Finding the minimum, **$M$

So far you’ve found:

Use the equations above to find the minimum $M$ in terms of $b$.

Almost; just be careful with the minus signs!

To see why $M=−2b $, substitute the above equations and simplify:

Finally, rearrange $b=−2M$ by dividing both sides by $−2$, getting $−2b =M$.

You just found something pretty powerful: given any quadratic equation $x_{2}+bx+c=0$, you can find its minimum with very little effort. Test it out: what value of $x$ minimizes $x_{2}−12x+42=0$?

Careful with the minus signs. You should have calculated:

Our trick lets us find the minimum without much analysis at all: just divide by $2$ and flip the sign!

**Finding the distance, **$d$

The minimum $M$ is right in the middle of the two roots, $L$ and $R$. To find those roots, we just need to know how far away they are from the middle $M$. That is, we need to find the distance $d$.

Here are some equations you’ve found so far:

Use these equations to find $c$ in terms of $b$ and $d$. You might want pencil and paper again ...

Not quite. Here’s the full calculation — it’s a bit long but mostly just plugging in numbers and simplifying:

Great job, that was a difficult one! 💪

Now to find the distance $d$, you just need to do a little bit of rearranging. Take the following equation for $c$, and rearrange it to get the distance $d$ in terms of $b$ and $c$. (Assume $d≥0$.)

You need to take the square root of $d$. Here’s the full rearrangement:

(Since we’re assuming $d≥0$, we can remove the $±$.)

This is another powerful tool in your toolbox! Given any quadratic equation $x_{2}+bx+c=0$, you can now quickly find the distance $d$ between its minimum $M$ and either root ($L$ or $R$). And $2d$ is the distance between the roots, $R$ minus $L$.

Try it out: for the equation $x_{2}−10x+16=0$, what’s the distance between its roots, i.e. $R−L$?

Almost!! It’s true that $d=3$, because:

The full distance between the roots, $R−L,$ is twice this, i.e. $2d=6$.

You’re very close now. Here are some equations you’ve found:

Use these to **find the restricted quadratic formula!** That is, find the roots $L$ and $R$ of any equation $x_{2}+bx+c=0$, in terms of $b$ and $c$ ...

🥁🥁🥁

Not quite. You just need to plug in the values of $M$ and $d$:

Great job! You found the quadratic formula! 🥳

You can now quickly solve any equation of the form $x_{2}+bx+c=0$, with no guessing or graphs! Wield your newfound power on this quadratic: what are the roots of $x_{2}+6x+5=0$?

Here’s the correct calculation:

So you’ve found the quadratic formula for a restricted form where $a=1$. But what about that general form, $Ax_{2}+Bx+C=0$?

It turns out that this is just a small modification to your formula!

**Generalizing to **$A=1$

Here was the original equation we wanted to solve:

In turns out you can convert this equation into the restricted form $x_{2}+bx+c=0$. To do so, divide both sides by $A$. With a little algebra, what do you get?

Here’s the correct calculation:

So we have:

For your final challenge, substitute $b$ and $c$ into your restricted quadratic equation, and **find the general quadratic formula!** After many simplification steps, what do you get?

Nice! It’s the mysterious formula from your old school textbook! 🥳

The correct simplification is:

With this lesson’s techniques, you can rediscover the quadratic formula if you ever forget it:

We can transform any quadratic like $Ax_{2}+Bx+C=0$ into a restricted quadratic like $x_{2}+bx+c=0$. Its graph is a symmetrical ‘bowl’ shape called a parabola. In general, this quadratic has two solutions, $L$ and $R$, and it can be rewritten as $(x−L)(x−R)=0$.

These roots both lie $d$ away from the middle point of the bowl, $M$. By multiplying out $(x−L)(x−R)$, and substituting $M$ and $d$, we find how the roots relate to $b$ and $c$. Some algebra and simplification lead naturally to the quadratic formula.

(This lesson was inspired by A new way to make quadratic equations easy.)