User:J D Bowen/Lecture

A derivation of the quadratic formula, or, how to solve $$ax^2+bx+c=0 \ $$ for $$x \ $$.

Why the average of the zeroes is $$-b/2a \ $$
We worked out before why the average of the zeroes of a quadratic is $$-b/2a \ $$, but let's do it again as a refresher:

Suppose you have some quadratic $$ax^2+bx+c \ $$, which factors as $$(px+q)(rx+s) \ $$.

If we distribute this out, note that we get $$ax^2+bx+c= prx^2+(ps+qr)x+qs \ $$, which implies $$a=pr, \ b= ps+qr \ $$.

Now note that the two zeroes of $$(px+q)(rx+s)=0 \ $$ will be the solutions of $$px+q=0, \ rx+s=0 \ $$. It should not be hard to check that these two solutions are $$x_1=-q/p, \ x_2=-s/r \ $$. Then the line of symmetry of the graph, which is the average of the two zeroes, is

$$\frac{x_1+x_2}{2} = \frac{\frac{-q}{p}+\frac{-s}{r}}{2} = \frac{1}{2}\left({ \frac{-q}{p}+\frac{-s}{r} }\right) = \frac{1}{2}\left({ \frac{-qr}{pr}+\frac{-sp}{rp} }\right) = \frac{-(qr+sp)}{2pr} \ $$.

But remember from our distribution that $$pr=a, \ ps+qr=b \ $$. So $$\frac{-(qr+sp)}{2pr} = \frac{-b}{2a} \ $$.

The Quadratic Formula
Since the average of the two zeroes $$x_1, \ x_2 \ $$ is $$-b/2a \ $$, it makes sense that they are the same distance from that average; that is, there is some number $$d \ $$ such that $$x_1=-b/2a -d, \ x_2 = -b/2a+d \ $$.

Given that both of these should be solutions, we write $$x=\tfrac{-b}{2a}\pm d \ $$ and plug this into our equation:

$$ax^2+bx+c = 0 \ $$

$$a\left({ \tfrac{-b}{2a}\pm d}\right)^2+b\left({\tfrac{-b}{2a}\pm d}\right)+c = 0 \ $$

$$a\underbrace{\left({ \tfrac{-b}{2a}\pm d}\right)\left({ \tfrac{-b}{2a}\pm d}\right)}_{\text{expanding the square}}+b\left({\tfrac{-b}{2a}\pm d}\right)+c = 0 \ $$

Now we distribute that expanded square; this is tricky, because of the $$\pm \ $$, but we can do it. Suppose we're examining the $$+ \ $$ case. Then that expanded square becomes

$$\left({ \tfrac{-b}{2a}+ d}\right)\left({ \tfrac{-b}{2a}+ d}\right)= \frac{(-b)(-b)}{(2a)(2a)} + d\frac{-b}{2a}+d\frac{-b}{2a}+d^2 = \frac{b^2}{4a^2} + 2d \frac{-b}{2a} +d^2 \ $$.

If we were examining the negative case, we'd have

$$\left({ \tfrac{-b}{2a}- d}\right)\left({ \tfrac{-b}{2a}- d}\right)= \frac{(-b)(-b)}{(2a)(2a)} - d\frac{-b}{2a}-d\frac{-b}{2a}+d^2 = \frac{b^2}{4a^2} - 2d \frac{-b}{2a} +d^2 \ $$.

And so for the general case, we can write

$$\left({ \tfrac{-b}{2a}\pm d}\right)^2 = \frac{b^2}{4a^2}\pm 2d\frac{-b}{2a} +d^2 \ $$

Therefore, our original equation becomes

$$a\underbrace{\left({\frac{b^2}{4a^2}\pm 2d\frac{-b}{2a} +d^2}\right)}_{\text{we distributed}}+b\left({\tfrac{-b}{2a}\pm d}\right)+c = 0 \ $$

We can bring that a into that set of parentheses:

$$\underbrace{\left({\frac{ab^2}{4a^2}\pm 2d\frac{-ab}{2a} +ad^2}\right)}_{\text{brought the a in}}+b\left({\tfrac{-b}{2a}\pm d}\right)+c = 0 \ $$

and start cancelling

$$\underbrace{\left({\frac{b^2}{4a}\pm (-bd) +ad^2}\right)}_{\text{cancelled} \ a}+b\left({\tfrac{-b}{2a}\pm d}\right)+c = 0 \ $$.

Similarly, we can bring the b inside the parentheses:

$$\left({\frac{b^2}{4a}\pm (-bd) +ad^2}\right)+\underbrace{\left({\tfrac{-b^2}{2a}\pm bd}\right)}_{\text{brought b in}}+c = 0 \ $$

and now we can remove our parentheses:

$$\frac{b^2}{4a}\pm (-bd) + ad^2 +\frac{-b^2}{2a}\pm bd+c = 0 \ $$.

Let's rearrange a few terms, and combine all our $$\pm \ $$ terms, and see if we can't simplify that a bit:

$$\underbrace{\frac{b^2}{4a}+\frac{-b^2}{2a}}_{\text{fractions together}}+ad^2+c \underbrace{\pm \left({ (-bd)+bd }\right)}_{\text{plus/minus terms together}} = 0 \ $$

Notice that now we have a pair of fractions on the far left we should probably add, and a zero sum in parentheses. So let's make their denominators the same on the left there:

$$\frac{b^2}{4a}+\underbrace{\frac{-2b^2}{4a}}_{\text{multiply top/bottom by 2}}+ad^2+c \pm \underbrace{0}_{\text{-bd+bd=0}} = 0 \ $$

Performing the additions, we get

$$\underbrace{\frac{-b^2}{4a}}_{b^2-2b^2=-b^2}+ad^2+c = 0 \ $$

Now we can start solving for $$d^2 \ $$ to get

$$ad^2+c=\frac{b^2}{4a} \ $$

$$ad^2 = \frac{b^2}{4a}-c \ $$

To perform the subtraction on the right, we must make the denominators the same:

$$ad^2 = \frac{b^2}{4a}-\underbrace{\frac{4ac}{4a}}_{\text{multiply top/bottom by 4a}} \ $$

$$ad^2 = \frac{b^2-4ac}{4a} \ $$

and divide by a to get

$$\frac{1}{a}ad^2 = \frac{b^2-4ac}{4a}\frac{1}{a} \ $$

or

$$d^2= \frac{b^2-4ac}{4a^2} \ $$.

Taking the square root of both sides, we have

$$d=\sqrt{\frac{b^2-4ac}{4a^2}}= \frac{\sqrt{b^2-4a}}{2a} \ $$.

Since we suspect the solutions to the equation $$ax^2+bx+c = 0 \ $$ is $$x=-b/2a\pm d \ $$, we should have

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

We will call this result the quadratic formula.