Engineering graduate here. how the hell do you derive the quadratic formula?

The only poster not trolling

>engineering
stfu

Mosties wanna derive solutions froms it nots tha forumular itselph

You need to use the Dunning-Kruger transformation to be able to do that

>we derived it
What do you mean by "we", Peasant?

The common response will be "complete the square". or words to that effect.

However, there is an alternate route which applies equally well as a starting-point to an algorithm for all general polynomials, where an (elementary) algebraic solution exists. That is, just up through the quartic. this alternate method is interesting in that it kicks off the first step of a general algo for all these cases.

The alternate route is: make a substitution of variable, plug'n'chug in a clever way, and then back-substitute! voila, same result by a different method. See pic related. (something I started writing up months ago but have left on pause for a while).

I had the pleasue of deriving the cubic and quartic equations for myself in previous Veeky Forums threads, which I consider to have been successes. I'd still like to go back and produce an article about same. I also noted this little commonality (and I'd like to study Galois Theory), but , y'know, life.

I seem to remember getting stuck on the notion of resolvent cubics or similar, in a bid to verify something for myself. Lots of tedium and autism that I actually want to go back over at some point, for the simple reason that "solving up through the quartic" and then expressing same as a flowchart, or algorithm, is a feasible and tantalizingly finite exercise. Just a lot of bookkeeping though.

I don't want to hear any smarty-pants bringing up Bring radicals, that's for later. :^)

I tried and now I am coughing up blood. I called 911 and they just laughed at me. They asked where I came up with the theta map in the Hodge theatre I defined, and I had no answer. What do I do? Am I dying?

Well I have this in my computer, lucky of you =D.

clutch af user

kill yourself

This way is better. It's not just a "clever substitution", the function is symmetric about x = b/2a.

meant this post

...

[math] \displaystyle
\\ ax^2 + bx +c = 0 \; \; \; \; \; \; | \; \cdot 4a \\
4a^2x^2 + 4abx + 4ac = 0 \\
4a^2x^2 + 4abx = -4ac \; \; \; \; \; \; | \; +b^2 \\
4a^2x^2 + 4abx +b^2 = b^2 -4ac \\
(2ax + b)^2 = b^2 -4ac \\
2ax + b = \pm \sqrt{b^2 - 4ac} \\
2ax = -b \pm \sqrt{b^2 - 4ac} \\
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
[/math]

>engineering graduate
civil engineering graduate, I assume

Why don't you try solving a polynomial where all the constants are unknowns?

Here ya go:

sites.google.com/site/butwhymath/algebra/the-quadratic-formula

Nice bait.

1/10

Nice.

it took GauB 40 years to prove and it's not completely proven yet

What are some of the most employable majors in STEM?

computer science and nothing else

Thank you for the compliment, and food for thought. Oddly at first glance I fail to recognize what you've referred to.

>Computer science
>NOTHING ELSE

FUCKING KEK

None of them will be enjoyable if you're a pleb who doesn't realize that's a subjective question, I'm passionate about chemistry so I find chemistry enjoyable,

true although Tao made some significant progress lately

He said employable

EE and ME as long as you are not an international student who is looking for some company to pay for their visa.

Sorry, that wasn't very clear.

If you have a quadratic function f(x) = a x^2 + bx + c then it attains its extremum where f'(x) = 2ax + b = 0, or x = -b/(2a). So, it attains its extremum at x = 0 exactly when b = 0. And when b = 0 the roots are easy to find: x = +/-sqrt(-c/a). So, for any other quadratic equation we can translate it horizontally so that it *does* attain its supremum at x = 0, and then we can solve the equation. (t - b/(2a)) is exactly what you need to substitute to do the translation.

Oh, I see. =D

I was thinking in high-school terms of a quadratic form, or parabola, if we take the graph as an even function (with the appropriate translation).

Indeed, ax^2 by itself presents no difficulty to see as an even function if we allow a to vary over the reals. Even adding the constant term " +c " with real c again poses no mental challenge - the graph, so to speak, is shifted up and down.

Where things get slightly interesting is with the addition of the linear term bx, but your adroit reply gets ahead of this. I suppose that the other way to put things is to ask under whether the general quadratic function f(x) = ax^2 + bx + c is always an even function "under the appropriate horizontal translation". Then, if so, what "horizontal translation of the graph" the general quadratic function can always take in order to arrive at an even function/symmetrical "graph". I further expect that if I were to actually do the straightforward exercise, I'd end in much the same place, but I haven't convinced myself of that as of yet as I'm doing something else to amuse myself tonight.

Try the pq formular!

A fully-qualified member of the homosapien race

Yes, it can always be even by a change of basis as that user described.

To shift a polynomial left by [math]\alpha[/math] you can expand [math]a(x-\alpha)^2+b(x-\alpha)+c[/math]. Since polynomial coefficients are linear combinations of powers of the variable, you can make a vector of the relevant powers of your variable, and polynomial translation left and right on the x-axis is encapsulated by matrix multiplication (row vector of coefficients times a square matrix). This fact can be put to interesting use for things like splines.