Have you ever worked out the solutions of a 4th degree polynomial using the quartic formula? By hand, not coding it into a computer.
If not, why don't you do it right now?
>>8168547
>why don't you do it right now?
Because it would offer zero enlightenment to do so.
>>8168554
It's good algebra practice and you can say you did it.
looks too hard
>>8168554
>good algebra practice
Maybe.
>you can say you did it.
I can't think of a single hypothetical individual who wouldn't find it laughable for someone to proclaim they did it. "le autism" etc.
>>8168561
>I can't think of a single hypothetical individual who wouldn't find it laughable for someone to proclaim they did it.
Screw my earlier post, I'm doing it now.
>>8168554
If you already know it.
>>8168547
>image for ants
>>8168677
While you're at it, want to give all of those repeated pieces symbols to reduce duplication in the full equation for me because I am le lazy? Maybe then I'd work it out.
this man is my husbando
>>8168685
He's pretty fuckin rad. It's a shame he seems to have committed suicide over a relationship. The duel shit seems pretty weak, I remember seeing a letter written in advance asking his friends not to blame the man who would kill him.
I'm too much of an undergrad to understand what he did but he's still one of my faves.
>>8168547
I haven't test-driven a specific quartic as OP suggests, but I have gone into detail about how to actually derive the quartic formula. I am the author/OP of the /sci/ thread archived here:
https://warosu.org/sci/thread/7613239
I did "test-drive" the cubic formula for specific polynomials, however, numerically.
At some point, I would like to revisit the this topic but I've been busy with other things. I gained one insight from the operation: that you can start off a solution for any of the degree 1-4 cases with a substitution of variable. That's what I want to write up (pic related, weaponized first draft autism in need of editing).
no but i can derive the cubic formula
>>8168707
Apparently the guy he was going to duel was that good a shooter. If I were him I'd fear that I wouldn't survive the duel either.
>>8168796
It's not so much that he knew he would die, but that it seemed he put himself in a situation to die intentionally and wanted his friends to hold no grudge. In any case, I haven't looked into it in quite a while, and for all I know I was reading bullshit to begin with.
>>8168799
That depends on whether love is intentional, and given that it's early 19th century France we're talking about I doubt it was in his case.
reminds me of pages long derivations for things like common mode gain of instrumentation op-amps.
very tedium, much bookkeeping <doge.jpg>
>>8168547
Why not complete the square and solve each quadratic term individually?
Once upon a time, I formatted the roots of the cubic into a "nice" form, kek:
"Let ax^3 + bx^2 + cx + d = 0. Then
[math] \displaystyle x = \bigg( - \frac{1}{3a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 + \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } + \bigg( - \frac{1}{3a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 - \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } - \frac{b}{3a} [/math]
[math] \displaystyle x = \bigg( \frac{ 1 + \sqrt{3} i }{6a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 + \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } + \bigg( \frac{ 1 - \sqrt{3} i }{6a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 - \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } - \frac{b}{3a} [/math]
[math] \displaystyle x = \bigg( \frac{ 1 - \sqrt{3} i }{6a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 + \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } + \bigg( \frac{ 1 + \sqrt{3} i }{6a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 - \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } - \frac{b}{3a} [/math]
."
Wiki has some caveats between whether the coefficients are real or complex, and I never closed the (personal-understanding) loop on that. In multiple places, there is a preference for a much uglier form with roots and such in the denominators. This worked fine for my spot-checks with real coefficients and complex variables. Something about permuting the roots.
>>8169408
cool quates bro
>>8168556
i dont need to practice algebra brainlet
And, this was how I rendered the quartic formula after following along with wiki. The former equation compares favorably with one line at wiki, I just insisted on using my own lettering scheme.
"Let ax^4 + bx^3 + cx^2 + dx + e = t^4 + pt^2 + qt + r = 0. Then
[math] \displaystyle x = \frac{ \pm_{1} \sqrt{ 2y+p } \pm_{2} \sqrt{ -(2y + 3p \pm_{1} \frac{q}{2 \sqrt{2y+p } } ) } }{2} - \frac{b}{4a} [/math]
or
[math] \displaystyle x = \pm_{1} \sqrt{ \frac{ -p \pm_{2} \sqrt{ p^2 - 4r } }{2} } - \frac{b}{4a} [/math] ."
The first equation is how I write an (original) quartic equation in its general form, while the next equation is the equivalent depressed polynomial, whose coefficients p,q and r are merely expressed algebraically in terms of a-e - notice the presence of p-r throughout the forms. y is a clever new variable of some sort which is introduced during the tall weeds part, and since a is manifestly nonzero, the only sticking point with the former equation is the situation when 2y + p = 0, which then entails the latter equation as a replacement. It appears that there are two repeated roots in this situation.
>>8169454
... or not ,since there are two distinct pm forms in the latter equation (potentially occasioning 4 roots and not two), one of which I did not notice while pasting pasta.
