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The quadratic formula: a tutorial

So you need to solve 2x2βˆ’3x=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 2x^2 - 3x = 2. You remember the word β€˜quadratic’ and something like 4acβˆ’b2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \sqrt{4ac - b^2} … or was it +b2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} +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! πŸ’ͺ

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\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \A, B\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \B and C\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \C:

Ax2+Bx+C=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \A x^2 + \B x + \C = 0

For example, 2x2βˆ’3x=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 2x^2 - 3x = 2 is a quadratic equation. What are its values of A\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \A, B\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \B and C\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \C?

Careful with the minus signs. Rearrange 2x2βˆ’3x=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 2x^2 - 3x = 2 into the form Ax2+Bx+C=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{A}x^2 + \blue{B}x + \green{C} = 0. You can do this by subtracting 2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 2 from both sides. You end up with 2x2+(βˆ’3)x+(βˆ’2)=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{2}x^2 +\blue{(-3)}x + \green{(-2)} = 0.

Good, you paid attention to the minus sign! 😁

β€˜Solving’ a quadratic equation means finding the values of x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x that make the equation equal on both sides. These special values of x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x are called the roots πŸ₯• of the equation. Consider again our quadratic equation, 2x2βˆ’3xβˆ’2=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{2}x^2 -\blue{3}x - \green{2} = 0. Which of these is not a root?

Actually, x=βˆ’12\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=-\tfrac{1}{2} is a root, as you can see by plugging it into the equation:

2x2βˆ’3xβˆ’2=2(βˆ’12)2βˆ’3(βˆ’12)βˆ’2=(2Γ—14)+32βˆ’2=12+32βˆ’42=1+3βˆ’42=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \red{2}x^2 -\blue{3}x - \green{2} &= \red{2}(-\tfrac12)^2 -\blue{3}(-\tfrac12) - \green{2} \\ &= (2 \times \tfrac14) +\tfrac32 - 2 \\ &= \tfrac12+\tfrac32 - \tfrac42 \\ &= \tfrac{1+3-4}2 \\ &= 0 \\ \end{align*}

Actually, x=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=2 is a root, as you can see by plugging it into the equation:

2x2βˆ’3xβˆ’2=2(2)2βˆ’3(2)βˆ’2=(2Γ—4)βˆ’6βˆ’2=8βˆ’8=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \red{2}x^2 -\blue{3}x - \green{2} &= \red{2}(2)^2 -\blue{3}(2) - \green{2} \\ &= (2 \times 4) -6 - 2 \\ &= 8-8 \\ &= 0 \\ \end{align*}

By plugging a value of x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x into a quadratic equation, you can quickly check whether that value is a root. Above, plugging in x=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=2 or x=βˆ’12\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=-\tfrac12 gives the valid equation 0=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 0=0, so they are both roots. But plugging in x=βˆ’2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=-2 gives the invalid equation 12=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 12=0, so it is not a root.

Unfortunately, trying random values of x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x is not an efficient way to find the roots!

Graphing the equation πŸ“ˆ

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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: βˆ’1x2βˆ’1x+2=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{-1}x^2 \blue{-1}x + \green{2} = 0. As we go, we’ll plot our guesses on a graph, like this:

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Here, we’re plotting y=βˆ’1x2βˆ’1x+2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y = \red{-1}x^2 \blue{-1}x + \green{2}. Our goal is to find an x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x where y=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y=0, as this would solve our equation. We’ve made a couple of guesses already at the ✘\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{✘} marks. What guesses have we made?

Not quite: those are the y\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y-values. We’ve made two guesses, x=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=0 and x=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=2, for which we calculated the y\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y-values y=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y=2 and y=βˆ’4\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y=-4. For x=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=0, we calculated y=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y=2:

y=βˆ’1x2βˆ’1x+2=βˆ’1(0)2βˆ’1(0)+2=0+0+2=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} y &= \red{-1}x^2 \blue{-1}x + \green{2} \\ &= \red{-1}(0)^2 \blue{-1}(0) + \green{2} \\ &= 0 + 0 + \green{2} \\ &= 2 \\ \end{align*}

In the same way, for the guess x=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=2, we calculated y=βˆ’4\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y=-4:

y=βˆ’1x2βˆ’1x+2=βˆ’1(2)2βˆ’1(2)+2=βˆ’4βˆ’2+2=βˆ’4\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} y &= \red{-1}x^2 \blue{-1}x + \green{2} \\ &= \red{-1}(2)^2 \blue{-1}(2) + \green{2} \\ &= -4 - 2 + \green{2} \\ &= -4 \\ \end{align*}

Right, we calculated two points: (x=0,y=2)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (x=0, y=2) and (x=2,y=βˆ’4)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (x=2, y=-4).

These two guesses were unsuccessful: they did not solve the equation βˆ’1x2βˆ’1x+2=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{-1}x^2 \blue{-1}x + \green{2} = 0. What would a successful guess β€” that is, a root β€” look like on our graph?

It’s the other way around. The y\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y-axis line is the set of points where x=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=0, so a ✘\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{✘} on the y\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y-axis would just be the guess x=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=0.

We’re trying to solve βˆ’1x2βˆ’1x+2=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{-1}x^2 \blue{-1}x + \green{2}=0. We’re plotting y=βˆ’1x2βˆ’1x+2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y = \red{-1}x^2 \blue{-1}x + \green{2}, so finding a point where y=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y=0 would solve our equation.

The x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x-axis line is the set of points where y=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y=0. So all ✘\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{✘} on the x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 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=βˆ’1x2βˆ’1x+2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y = \red{-1}x^2 \blue{-1}x + \green{2} for the rest of the range βˆ’5≀x≀5\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} -5 \leq x \leq 5. Which of the following curves does it look like? (You might want to grab paper and pencil.)

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It’s not curve B\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{\bold B}. Try for example x=βˆ’3\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=-3, where you should get y=βˆ’4\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y=-4. This is far from the point (x=βˆ’3,y=4)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (x=-3,y=4) on curve B\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{\bold B}. If you plot all points correctly, you’ll find they follow curve C\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{\bold C} above.

It’s not curve A\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{\bold A}. Try for example x=5\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=5, where you should get y=βˆ’28\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y=-28. This is far from the point (x=5,y=βˆ’3)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (x=5, y=-3) on curve A\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{\bold A}. If you plot all points correctly, you’ll find they follow curve C\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{\bold C} above.

After plotting curve C\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{\bold C}, its roots should be clear. What are they?

No, that’s the intersection of C\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{C} with the y\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y-axis. We want the intersection of C\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{C} with the x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x-axis. These intersections are at x=βˆ’2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=-2 and x=1\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=1.

That’s one root, but there’s another at (x=βˆ’2,y=0)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (x=-2, y=0).

Notice that curve C\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{\bold C} follows an upside-down β€˜bowl’ shape πŸ₯£. If you plot some more quadratics, you’ll find that they all follow a similar β€˜bowl’ or β€˜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 β€˜bowl’ shape, called a parabola.

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

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No, that could actually be a quadratic! Only D\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{\bold D} is definitely not quadratic.

Well done. D\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{\bold D} could not be a quadratic, because it has an β€˜S’ shape.

A\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{\bold A} and C\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{\bold C} are both parabolas (β€˜bowl’ shapes, where bowl A\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{\bold A} happens to be upside-down). B\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{B} could also be part of a very stretched parabola. But the curve D\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{\bold D} has an β€˜S’ shape, so it could not be quadratic.

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

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

No, this could not work. Only a β€˜U’ shape could fit through these three points, and the only x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x-value that looks like it could work is x=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=2:

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

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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

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Here’s a simpler equation with two unknowns, P\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{P} and Q\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{Q}:

PΓ—Q=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{P} \times \pink{Q}=0

What best describes all possible solutions to this equation?

This does satisfy PΓ—Q=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} P \times Q=0, but there are more solutions! For example, P=5\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} P=5 and Q=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} Q=0. You can visualize all the possible solutions by writing them out in a big table:

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The result is zero all down the P=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} P = 0 column and all along the Q=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 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\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} and R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R}:

(xβˆ’L)(xβˆ’R)=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{(x-L)}\pink{(x-R)}=0

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=xβˆ’L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{P}=\orange{x-L} and Q=xβˆ’R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{Q}=\pink{x-R}. If P=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{P}=0, then xβˆ’L=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{x-L}=0. If Q=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{Q} = 0, xβˆ’R=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{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 xβˆ’R=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x-\pink{R}=0 and add R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R} to both sides. You should get x=R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=\pink{R}.

Multiplying out

The equation (xβˆ’L)(xβˆ’R)=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{(x-L)}\pink{(x-R)}=0 is actually a quadratic! You might remember how to β€˜multiply out’ brackets, like this:

(a+b)(c+d)=ac+ad+bc+bd\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (a+b)(c+d) = ac + ad + bc + bd

Using this method, multiply out (xβˆ’L)(xβˆ’R)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (x-\orange{L})(x-\pink{R}).

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

a=xb=βˆ’Lc=xd=βˆ’R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} a &= x \\ b &= -\orange{L} \\ c &= x \\ d &= -\pink{R} \\ \end{align*}

Multiplying out, you should get

(xβˆ’L)(xβˆ’R)=xx+x(βˆ’R)+(βˆ’L)x+(βˆ’L)(βˆ’R)=x2+(βˆ’Lβˆ’R)x+LR=x2βˆ’(L+R)x+LR\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} (x-\orange{L})(x-\pink{R}) &= xx + x(-\pink{R}) + (-\orange{L})x + (-\orange{L})(-\pink{R}) \\ &= x^2 +(-\orange{L} -\pink{R})x + \orange{L}\pink{R} \\ &= x^2 -(\orange{L}+\pink{R})x + \orange{L}\pink{R} \end{align*}

Right. The negatives cancel when multiplying, and we can factor a βˆ’x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} -x out of the two middle terms.

So if I gave you this equation:

x2βˆ’(L+R)x+LR=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2-(\orange{L}+\pink{R})x+\orange{L}\pink{R} = 0

What are its roots? That is, what values of x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x solve this equation?

No, it’s simpler than that! Earlier you found the solutions to (xβˆ’L)(xβˆ’R)=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (x-\orange{L})(x-\pink{R})=0 are x=L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=\orange{L} or x=R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=\pink{R}. And you found that (xβˆ’L)(xβˆ’R)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (x-\orange{L})(x-\pink{R}) can be rearranged as x2βˆ’(L+R)x+LR\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2-(\orange{L}+\pink{R})x+\orange{L}\pink{R}. Rearranging an equation doesn’t change its solutions, so the solutions are still the same: x=L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=\orange{L} or x=R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=\pink{R}.

Still don’t believe me? Plug in x=L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x = \orange{L} and verify:

x2βˆ’(L+R)x+LR=0L2βˆ’(L+R)L+LR=0L2βˆ’(L2+RL)+LR=0L2βˆ’L2βˆ’RL+LR=00=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} x^2-(\orange{L}+\pink{R})x+\orange{L}\pink{R} = 0 \\ \orange{L}^2-(\orange{L}+\pink{R})\orange{L}+\orange{L}\pink{R} = 0 \\ \orange{L}^2-(\orange{L}^2+\pink{R}\orange{L})+\orange{L}\pink{R} = 0 \\ \orange{L}^2-\orange{L}^2-\pink{R}\orange{L}+\orange{L}\pink{R} = 0 \\ 0 = 0 \end{align*}

You can check the same with x=R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=\pink{R}.

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

Hopefully, the above equation looks familiar …

Ax2+Bx+C=0x2βˆ’(L+R)x+LR=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \red{A}x^2 + \blue{B}x + \green{C} = 0 \\ x^2-(\orange{L}+\pink{R})x+\orange{L}\pink{R} = 0 \end{align*}

Yes, it’s a quadratic equation! But what are its values of A\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{A}, B\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \blue{B} and C\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{C}?

Almost. Just pay attention to the minus signs: B=βˆ’(L+R)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \blue{B}=-(\orange{L}+\pink{R}).

This is almost a general quadratic equation, except that A=1\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{A}=1. For now, we’re going to deal with quadratic equations of the more restricted form:

x2+bx+c=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2 + \blue{b}x + \green{c} = 0

The lowercase b\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \blue{b} and c\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{c} signal this restricted form where A=1\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{A}=1. Later, we’ll see how to deal with different values of A\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{A}.

If we graph the equation y=x2βˆ’(L+R)x+LR\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y = x^2-(\orange{L}+\pink{R})x+\orange{L}\pink{R}, which of the following could it look like?

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No, it can’t be curve B\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{\bold B} because that has no solutions β€” but we found earlier that (L+R)x+LR=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (\orange{L}+\pink{R})x+\orange{L}\pink{R} = 0 must have two solutions, x=L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=\orange{L} and x=R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=\pink{R}.

It can’t be curve A\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{\bold A} because inverted β€˜U’ shapes only happen with A<0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{A} < 0. To see why, think about extreme values, like x=Β±1000\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=\pm 1000. Out here, x2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2 becomes huge and the other terms become irrelevant.

Curve C\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{\bold C} is the only possible curve for a quadratic with A=1\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{A}=1, because it’s the only upwards β€˜bowl’ curve that crosses the x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x-axis. The curve of y=x2βˆ’(L+R)x+LR\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y = x^2-(\orange{L}+\pink{R})x+\orange{L}\pink{R} must follow a parabola, crossing the x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x-axis at the Left\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{\text{L}}\text{eft} and Right\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{\text{R}}\text{ight} root:

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Finding the minimum πŸ“‰

We’re almost there! Instead of finding L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} and R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R} directly, we’re first going to find the value exactly between them. Now consider the following graph of y=(xβˆ’1)(xβˆ’3)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y=(x-\orange{1})(x-\pink{3}):

Could not load <object> SVG

By now, you know how to find x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x values that make y=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y=0. Above, those solutions are x=L=1\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x = \orange{L} = 1 and x=R=3\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x = \pink{R} = 3.

But now a different question: in the graph above, what value of x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x above minimizes the value of y\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y?

Maybe you had the right idea looking at the point (x=2,y=βˆ’1)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (x=2,y=-1), where the bottom of the β€˜U’ shape is. However, I was looking for the x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x coordinate, x=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=2.

No, that minimizes x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x, but certainly doesn’t minimize y\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y. The y\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} y value becomes larger as the x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x value goes towards βˆ’βˆž\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} -\infty.

We’ll call this minimum x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x value M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M}. We can find M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} just by looking at the graph above. First we find the bottom point of the β€˜bowl’, at (x=2,y=βˆ’1)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (x=2,y=-1). Then M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} is its x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x coordinate, so M=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M}=2:

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This curve is beautifully symmetrical, isn’t it? It’s an exact mirror image along the x=M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x=\purple{M} curve, with the roots L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} and R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R} on either side. This symmetry will be the key to unlocking the quadratic formula!

Averages πŸ””

Because the curve is symmetrical, the minimum M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} is exactly between L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} and R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R}. That is, M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} is the mean (or average) of L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} and R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R}.

As a reminder, the mean M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} of any two numbers L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} and R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R} is defined as

M=L+R2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} = \frac{\orange{L}+\pink{R}}2

For example, if L=3\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} = 3 and R=7\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R} = 7, what is their mean M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M}?

You’re one off. Try a calculator: 3+72=5\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \frac{3 + 7}2 = 5.

5\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 5 is exactly in the middle of 3\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 3 and 7\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 7, as we can visualize on a number line:

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That distance d=2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d}=2 is the same on both sides. (That’s what it means for M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} to be β€˜in the middle’!)

What’s interesting is that we can define these the other way around. Say M=4\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M}=4 and d=6\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d}=6. Then what are L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} and R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R}? (Assume that L≀R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} \leq \pink{R}.)

Not quite; you probably got M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} and d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d} the wrong way around. Here’s the correct calculation:

L=Mβˆ’d=4βˆ’6=βˆ’2R=M+d=4+6=10\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \orange{L} &= \purple{M}-\green{d} = 4-6 = -2 \\ \pink{R} &= \purple{M}+\green{d} = 4+6 = 10 \end{align*}

In general, we can write:

L=Mβˆ’dR=M+d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \orange{L} &= \purple{M}-\green{d} \\ \pink{R} &= \purple{M}+\green{d} \\ \end{align*}

Remember that our goal is to find L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} and R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R}, the roots of our quadratic equation, x2+bx+c=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2 + \blue{b}x + \green{c} = 0. But our strategy will be to first find M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} and d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d}, then plug those into the equations above to find the roots L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} and R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R}.

Finding the minimum, M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \bold{\purple{M}}

So far you’ve found:

b=βˆ’(L+R)L=Mβˆ’dR=M+d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \blue{b} &=-(\orange{L}+\pink{R}) \\ \orange{L} &= \purple{M}-\green{d} \\ \pink{R} &= \purple{M}+\green{d} \\ \end{align*}

Use the equations above to find the minimum M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} in terms of b\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \blue{b}.

Almost; just be careful with the minus signs!

To see why M=βˆ’b2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M}=-\frac{\blue{b}}2, substitute the above equations and simplify:

b=βˆ’(L+R)=βˆ’((Mβˆ’d)+(M+d))=βˆ’(M+Mβˆ’d+d)=βˆ’2M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \blue{b} &=-(\orange{L}+\pink{R}) \\ &= -((\purple{M}-\green{d}) +(\purple{M}+\green{d})) \\ &= -(\purple{M}+\purple{M} -\green{d}+\green{d}) \\ &= -2\purple{M} \end{align*}

Finally, rearrange b=βˆ’2M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \blue{b}= -2\purple{M} by dividing both sides by βˆ’2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} -2, getting βˆ’b2=M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} -\frac{\blue{b}}2 = \purple{M}.

You just found something pretty powerful: given any quadratic equation x2+bx+c=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2 + \blue{b}x + \green{c} = 0, you can find its minimum with very little effort. Test it out: what value of x\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x minimizes x2βˆ’12x+42=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2 \blue{-12}x + \green{42} = 0?

Careful with the minus signs. You should have calculated:

M=βˆ’b2=βˆ’βˆ’122=122=6\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} = -\frac{\blue{b}}2 = -\frac{\blue{-12}}2 = \frac{12}2 = 6

Our trick lets us find the minimum without much analysis at all: just divide by 2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 2 and flip the sign!

Finding the distance, d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \bold{\green{d}}

The minimum M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} is right in the middle of the two roots, L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} and R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R}. To find those roots, we just need to know how far away they are from the middle M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M}. That is, we need to find the distance d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d}.

Here are some equations you’ve found so far:

c=LRL=Mβˆ’dR=M+dM=βˆ’b2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \green{c} &= \orange{L}\pink{R} \\ \orange{L} &= \purple{M}-\green{d} \\ \pink{R} &= \purple{M}+\green{d} \\ \purple{M} &=-\tfrac{\blue{b}}2 \end{align*}

Use these equations to find c\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{c} in terms of b\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \blue{b} and d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d}. You might want pencil and paper again ...

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

c=LR=(Mβˆ’d)(M+d)=M2βˆ’Md+Mdβˆ’d2=M2βˆ’d2=(βˆ’b2)2βˆ’d2=b24βˆ’d2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \green{c} &= \orange{L}\pink{R} \\ &= (\purple{M}-\green{d})(\purple{M}+\green{d}) \\ &= \purple{M}^2 -\purple{M}\green{d} +\purple{M}\green{d} - \green{d}^2 \\ &= \purple{M}^2 - \green{d}^2 \\ &= (-\tfrac{\blue{b}}2)^2 - \green{d}^2 \\ &= \tfrac{\blue{b}^2}4 - \green{d}^2 \\ \end{align*}

Great job, that was a difficult one! πŸ’ͺ

Now to find the distance d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d}, you just need to do a little bit of rearranging. Take the following equation for c\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{c}, and rearrange it to get the distance d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d} in terms of b\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \blue{b} and c\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{c}. (Assume dβ‰₯0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d}\geq0.)

c=b24βˆ’d2\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{c} = \tfrac{\blue{b}^2}4 - \green{d}^2

You need to take the square root of d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d}. Here’s the full rearrangement:

c=b24βˆ’d2c+d2=b24d2=b24βˆ’cd=Β±b24βˆ’c\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \green{c} &= \tfrac{\blue{b}^2}4 - \green{d}^2 \\ \green{c} + \green{d}^2&= \tfrac{\blue{b}^2}4 \\ \green{d}^2&= \tfrac{\blue{b}^2}4 - \green{c} \\ \green{d} &= \pm\sqrt{\tfrac{\blue{b}^2}4 - \green{c}} \\ \end{align*}

(Since we’re assuming dβ‰₯0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d}\geq0, we can remove the Β±\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pm.)

This is another powerful tool in your toolbox! Given any quadratic equation x2+bx+c=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2 + \blue{b}x + \green{c} = 0, you can now quickly find the distance d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d} between its minimum M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} and either root (L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} or R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R}). And 2d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 2\green{d} is the distance between the roots, R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R} minus L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L}.

Try it out: for the equation x2βˆ’10x+16=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2 \blue{-10}x + \green{16} = 0, what’s the distance between its roots, i.e. Rβˆ’L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R}-\orange{L}?

Almost!! It’s true that d=3\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d}=3, because:

d=b24βˆ’c=(βˆ’10)24βˆ’16=1004βˆ’16=25βˆ’16=9=3\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \green{d}&= \sqrt{\tfrac{\blue{b}^2}4 - \green{c}} \\ &= \sqrt{\tfrac{(\blue{-10})^2}4 - \green{16}} \\ &= \sqrt{\tfrac{100}4 - \green{16}} \\ &= \sqrt{25 - \green{16}} \\ &= \sqrt{9} \\ &= 3 \\ \end{align*}

The full distance between the roots, Rβˆ’L,\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R}-\orange{L}, is twice this, i.e. 2d=6\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} 2\green{d}=6.

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

TheΒ rootsΒ LΒ andΒ R=MΒ±dM=βˆ’b2d=b24βˆ’c\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \text{The roots } \orange{L} \text{ and } \pink{R} &= \purple{M}\pm\green{d} \\ \purple{M} &=-\tfrac{\blue{b}}2 \\ \green{d}&= \sqrt{\tfrac{\blue{b}^2}4 - \green{c}} \end{align*}

Use these to find the restricted quadratic formula! That is, find the roots L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} and R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R} of any equation x2+bx+c=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2 + \blue{b}x + \green{c} = 0, in terms of b\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \blue{b} and c\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{c} ...


Not quite. You just need to plug in the values of M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} and d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d}:

TheΒ rootsΒ LΒ andΒ R=MΒ±d=(βˆ’b2)Β±(b24βˆ’c)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \text{The roots } \orange{L} \text{ and } \pink{R} &= \purple{M}\pm\green{d} \\ &=(-\tfrac{\blue{b}}2) \pm (\sqrt{\tfrac{\blue{b}^2}4 - \green{c}}) \end{align*}

Great job! You found the quadratic formula! πŸ₯³

You can now quickly solve any equation of the form x2+bx+c=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2 + \blue{b}x + \green{c} = 0, with no guessing or graphs! Wield your newfound power on this quadratic: what are the roots of x2+6x+5=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2 + \blue{6}x + \green{5} = 0?

Here’s the correct calculation:

TheΒ rootsΒ LΒ andΒ R=βˆ’b2Β±b24βˆ’c=βˆ’62Β±624βˆ’5=βˆ’3Β±9βˆ’5=βˆ’3Β±4=βˆ’3Β±2=βˆ’5Β orΒ βˆ’1\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \text{The roots } \orange{L} \text{ and } \pink{R} &= -\tfrac{\blue{b}}2 \pm \sqrt{\tfrac{\blue{b}^2}4 - \green{c}} \\ &= -\tfrac{\blue{6}}2 \pm \sqrt{\tfrac{\blue{6}^2}4 - \green{5}} \\ &= -3 \pm \sqrt{9 - \green{5}} \\ &= -3 \pm \sqrt{4} \\ &= -3 \pm 2 \\ &= {-5} \text{ or } {-1} \end{align*}

So you’ve found the quadratic formula for a restricted form where a=1\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{a}=1. But what about that general form, Ax2+Bx+C=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{A}x^2 + \blue{B}x + \green{C} = 0?

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

Generalizing to A≠1\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \bold{\red{A}\neq1}

Here was the original equation we wanted to solve:

Ax2+Bx+C=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{A}x^2 + \blue{B}x + \green{C} = 0

In turns out you can convert this equation into the restricted form x2+bx+c=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2 + \blue{b}x + \green{c} = 0. To do so, divide both sides by A\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{A}. With a little algebra, what do you get?

Here’s the correct calculation:

Ax2+Bx+C=0Ax2+Bx+CA=0AAx2A+BxA+CA=0x2+BAx+CA=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \red{A}x^2 + \blue{B}x + \green{C} &= 0 \\ \frac{\red{A}x^2 + \blue{B}x + \green{C}}{\red{A}} &= \frac{0}{\red{A}} \\ \frac{\red{A}x^2}{\red{A}} + \frac{\blue{B}x}{\red{A}} + \frac{\green{C}}{\red{A}} &= 0 \\ x^2 + \tfrac{\blue{B}}{\red{A}}x + \tfrac{\green{C}}{\red{A}} &= 0 \\ \end{align*}

So we have:

TheΒ rootsΒ LΒ andΒ R=βˆ’b2Β±b24βˆ’cb=BAc=CA\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \text{The roots } \orange{L} \text{ and } \pink{R} &= -\frac{\blue{b}}2\pm\sqrt{\frac{\blue{b}^2}4 - \green{c}} \\ \blue{b} &= \tfrac{\blue{B}}{\red{A}} \\ \green{c} &= \tfrac{\green{C}}{\red{A}} \\ \end{align*}

For your final challenge, substitute b\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \blue{b} and c\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{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! πŸ₯³

The correct simplification is:

LΒ andΒ R=βˆ’b2Β±b24βˆ’c=βˆ’(BA)2Β±(BA)24βˆ’CA=βˆ’B2AΒ±B24A2βˆ’CA=βˆ’B2AΒ±B24A2βˆ’4CA4A2=βˆ’B2AΒ±B2βˆ’4AC4A2=βˆ’B2AΒ±B2βˆ’4AC4A2=βˆ’B2AΒ±B2βˆ’4AC2A=βˆ’BΒ±B2βˆ’4AC2A\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \begin{align*} \orange{L} \text{ and } \pink{R} &= -\frac{\blue{b}}2\pm\sqrt{\frac{\blue{b}^2}4 - \green{c}} \\ &= -\frac{(\frac{\blue{B}}{\red{A}})}2\pm\sqrt{\frac{(\frac{\blue{B}}{\red{A}})^2}4 - \frac{\green{C}}{\red{A}}} \\ &= -\frac{\blue{B}}{2\red{A}} \pm \sqrt{\frac{\blue{B}^2}{4\red{A}^2} - \frac{\green{C}}{\red{A}}} \\ &= -\frac{\blue{B}}{2\red{A}} \pm \sqrt{\frac{\blue{B}^2}{4\red{A}^2} - \frac{4\green{C}\red{A}}{4\red{A}^2}} \\ &= -\frac{\blue{B}}{2\red{A}} \pm \sqrt{\frac{\blue{B}^2 - 4\red{A}\green{C}}{4\red{A}^2}} \\ &= -\frac{\blue{B}}{2\red{A}} \pm \frac{\sqrt{\blue{B}^2 - 4\red{A}\green{C}}}{\sqrt{4\red{A}^2}} \\ &= -\frac{\blue{B}}{2\red{A}} \pm \frac{\sqrt{\blue{B}^2 - 4\red{A}\green{C}}}{2\red{A}} \\ &= \frac{-\blue{B} \pm \sqrt{\blue{B}^2 - 4\red{A}\green{C}}}{2\red{A}} \\ \end{align*}

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

We can transform any quadratic like Ax2+Bx+C=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \red{A}x^2 + \blue{B}x + \green{C} = 0 into a restricted quadratic like x2+bx+c=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} x^2 + \blue{b}x + \green{c} = 0. Its graph is a symmetrical β€˜bowl’ shape called a parabola. In general, this quadratic has two solutions, L\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \orange{L} and R\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \pink{R}, and it can be rewritten as (xβˆ’L)(xβˆ’R)=0\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (x-\orange{L})(x-\pink{R})=0.

These roots both lie d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d} away from the middle point of the bowl, M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M}. By multiplying out (xβˆ’L)(xβˆ’R)\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} (x-\orange{L})(x-\pink{R}), and substituting M\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \purple{M} and d\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{d}, we find how the roots relate to b\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \blue{b} and c\def\A{\red{A}} \def\B{\blue{B}} \def\C{\green{C}} \green{c}. Some algebra and simplification lead naturally to the quadratic formula.

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