Having discussed the motivation of why something like a “quadratic formula” is a useful thing to discover and understand, I’d like to work through some of the ideas that might lead one to discover a quadratic formula.

Reducing the Number of Unknowns

Remember that the equation we care about is $ax^2 + bx + c = 0$, with $a \not = 0$ (so that this equation actually has $x^2$ as part of it). Imagine for a moment that we already know the correct value of $x$, so $x$ is a number rather than a variable. Dividing both sides of an equation by a number different from zero is always allowed, so we are allowed to divide by $a.$ This leads to the new equation

$x^2 + \dfrac{b}{a} x + \dfrac{c}{a} = 0.$

What does this tell us? Well, morally this means that we can ‘bake $a$‘ into $b$ and $c$. The logic is similar to the idea that if we know that $6 + 10 = 16$, then $3 + 5 = 8$ since dividing 6, 10, and 16 by two results in 3, 5, and 8. To make things a little more simple, we can define new variables $B = \dfrac{b}{a}$ and $C = \dfrac{c}{a}$, so that we really only have to solve

$x^2 + Bx + C = 0.$

If we want a value of $a$ other than 1, we can fix this by multiplying everything by the value of $a$ that we want.

Working Backwards and What it Teaches Us

We have already discussed a way to simplify our problem into the problem of factoring $x^2 + bx + c$ for some numbers $b$ and $c$. As is often helpful in all areas of problem solving, we might wonder whether the solutions might tell us anything about where we started. In fact, in this case they do. As an example, suppose that we already know in advance that the equation we care about has the solutions $x = 2$ and $x = 3$. This means that $2^2 + 2b + c = 0$ and $3^2 + 3b + c = 0$. Solving this is a bit tedious. However, we might remember from algebra class that quadratic equations have factorizations and that these factorizations look like

$x^2 + bx + c = (x - s)(x - t)$

where $x = s$ and $x = t$ are the solutions. We might also remember the distributive law of multiplication, that which tells us that $A(B - C) = AB - AC$ no matter what $A, B, C$ might be. Using this law with $A = x - s$, $B = x$, and $C = t$, we can conclude that

$(x - s)(x - t) = (x-s)x - (x-s)t,$

and using a very similar process we can conclude that

$(x - s)(x - t) = (x - s)x - (x - s)t = x^2 - sx - tx + st = x^2 - (s + t)x + st.$

We now go back for a moment to polynomials. We have assumed that $x^2 + bx + c$ is our initial polynomial, and that we can factor this polynomial as $(x - s)(x - t)$. We then used the distributive law to simplify this expression. If we assume that $x^2 + bx + c = (x - s)(x - t)$, then we have to conclude that

$x^2 + bx + c = x^2 - (s + t)x + st,$

from which we can conclude that $b = -(s+t)$ and $st = c$. We can now see that the solutions to $x^2 + bx + c = 0$ have a lot to do with $b$ and $c$. Since these things are related by mathematics, the first thing you might hope to do is to find an equation for $s$ and $t$. In our situation, we can in fact use a solution in this way. However, before the broadest solution is possible, it makes more sense to think about what turns out to be the easiest solution first. In mathematics, we call this situation a ‘perfect square.’ After solving the perfect square, the full solution is actually much easier.

Solving the Perfect Square

The idea of the perfect square is common throughout mathematics. The use of the term ‘square’ refers to the idea of the geometric dimension of the square, which is 2. The area of a square looks like $s^2$ where $s$ is the side length of the square. The use of the word ‘perfect’ refers essentially to the fact that squaring is the only idea going on – that is, if you know how to find $s^2$, then you don’t need anything else.

In terms of the algebra earlier, a ‘perfect square’ actually has identical roots. In other words, we actually learn that in fact $s = t$. Using this equality, we may deduce that

$(x - s)^2 = x^2 - 2sx + s^2.$

Our earlier discussion showed how to learn about the roots of $x^2 + bx + c$ by using $b$ and $c$. Namely, if $s$, $t$ are the two roots, then $b = -(s+t)$ and $c = st$. For the perfect square case, this becomes $b = - 2s$ and $c = s^2$.

To determine whether the we have a quadratic that is a perfect square, we first ask whether $c = s^2$ for some value of $s$. If it is, then we check whether it is possible to chose $s$ in such a way that $b = - 2s$. If so, then $x^2 + bx + c = (x - s)^2$, and the only solution to $x^2 + bx + c = 0$ is $x = s$.

From all of this, the most important thing to learn is that this perfect square format is fairly easy to deal with. We might wonder then if we can find a way to manipulate harder cases into something similar to perfect squares. It turns out that this is exactly the right way to go.

Reducing All Equations to Perfect Squares

We can now find a method of converting any equation to a perfect square case. To do this, we at first only need to know what $b$ is. If $x^2 + bx + c$ is a perfect square, then $b = -2s$ and $c = s^2$. The first of these equations means that $s = - \dfrac{b}{2}$, and therefore $c = s^2 = \dfrac{b^2}{4}$. Now, to make $x^2 + bx + c = 0$ easier to solve, we subtract $c$ from both sides and then add $\dfrac{b^2}{4}$ to both sides, which gives us

$x^2 + bx + \dfrac{b^2}{4} = \dfrac{b^2}{4} - c.$

Since the left hand side is now a perfect square with the value $s = - \dfrac{b}{2}$, and so we conclude that

$\bigg( x + \dfrac{b}{2} \bigg)^2 = \dfrac{b^2}{4} - c.$

To make this a little easier to deal with, we can multiply everything by 4 to get rid of the fractions. When we simplify this, we obtain

$(2x + b)^2 = b^2 - 4c.$

Now, we have our perfect square. We could try using something like the previous approach to perfect squares, but the fact that the right hand side isn’t 0 any more means that won’t work. What does work, however, is taking square roots. If we do this, we find out that

$2x + b = \pm \sqrt{b^2 - 4c},$

and solving for $x$ leads to the conclusion that

$x = \dfrac{-b \pm \sqrt{b^2 - 4c}}{2}.$

Final Step To the Quadratic Formula

We now have a quadratic formula for equations like $x^2 + bx + c = 0$. But we’d like a little bit more than this – we want to solve $ax^2 + bx + c = 0$. From the very beginning of the discussion, however, remember that we used the equation $x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$ in place of $ax^2 + bx + c = 0$. This means that to solve $ax^2 + bx + c = 0$, all we need to do is replace all the $b$‘s and $c$‘s in the previous formula with $\dfrac{b}{a}$ and $\dfrac{c}{a}$. When we do this and simplify, we finally arrive at the famous quadratic formula,

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$