Quadratic formula made simple as fuck.

So I know, it’s been a minute since I last posted.  I was travelling the world and shit; I ain’t got time to be teaching when I’m kicking back cheese, chocolate, and beer all day long.  Back to business.

Quadratic Formula

I don’t know how so many people memorize this goddamn monstrosity, but they do, and they likely don’t know where the fuck it came from.  Fuck that.  I’m going to show y’all where it comes from, building on my last post about completing the square.  You should be excited about this.

The quadratic formula is used to find the solutions to quadratic equations.  No shit, right?  It looks like this, and I hate it:

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

 Alright so what is a quadratic equation?  Anything that looks like this:

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

Graphically, that $ax^2+bx+c$ takes on a parabolic shape, like so:

Alright so what does it mean to find solutions for quadratic equations?  A solution is any value for x (or whatever fucking letter/s you are using) that makes the equation true.  So if we use the expression we graphed above, our quadratic equation would be:

$$2x^2-3x+1=0$$

So then, a solution to this equation would be any value for x that would make the left side of the equation equal to zero.  Another way to think about a solution is any point where the graph of the equation crosses the x-axis, since those are the places where the function is equal to zero.  This is also called finding zeroes, or roots, since throwing in fucking ambiguous language always helps.

So we know that the quadratic formula gives us the solutions to quadratic equations, but how the hell does it do that?  Because, you eager motherfucker, all the quadratic formula is is a factored version of the same damn equation.  All it is is a nasty-ass shortcut.  Watch me factor this shit.  I complete the square to do this, so if you ain’t up to speed on that, go learn something here.

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

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

$$[x^2+\dfrac{b}{a}x+(\dfrac{b}{2a})^2]+\dfrac{c}{a}=(\dfrac{b}{2a})^2$$

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

$$(x+\dfrac{b}{2a})=\pm\sqrt{\dfrac{b^2}{4a^2}-\dfrac{c}{a}}$$

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

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

For fuck’s sake!  I’d like to take this moment to say that deriving this shit is ugly as hell (uglier than I remembered it being) and I am going to try my damnest not to have to type anything like this clusterfuck out ever again.  Anyway…

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

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

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

Fucking done.  So you can see that the quadratic formula is just a mangled-beyond-belief version of the original equation.  So that means whatever we find for x using this formula is automatically a solution for the equation.  You’ll notice that there’s a $\pm$ in there, which is a result of taking the square root during the derivation of the formula.  This means that we may get two solutions – one using the “plus” of the $\pm\,$, and one using the “minus.”  This isn’t a problem, though, because the graph above shows that the graph crosses the x-axis twice – therefore there are two solutions.

So let’s find the solutions for the original equation and be done.  From before:

$$a=2,\,b=-3,\,c=1$$

So now plug them shits in and see what we get.

$$x=\dfrac{3\pm\sqrt{-3^2-4\times2\times1}}{2\times2}$$

$$x=\dfrac{3\pm\sqrt{9-8}}{4}$$

$$x=\dfrac{3\pm\sqrt{1}}{4}$$

$$x=\dfrac{3+\sqrt{1}}{4},\,x=\dfrac{3-\sqrt{1}}{4}$$

$$x=\dfrac{4}{4},\,x=\dfrac{2}{4}$$

$$x=1,\,x=0.5$$

I’ll be damned, that worked out nicely.  Compare those answers with our graph, and they seem to check out, so I say we’re done with this shit.

The Quadratic Formula.  A necessary evil.