The most fun way of solving this quartic equation

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What's funny is that you can actually rearrange this equation into x^4 = (x+1)^2. Then you take the square root of both sides and consider the two quadratics x^2 = ±(x+1), which are the same if you had used the quadratic formula. Idk if something like this is possible whenever the discriminant of the first "quadratic" simplifies nicely, but would be interesting to investigate. Edit: I just realized you could also write the equation as x^4 - (x+1)^2 = 0 and apply difference of squares to get (x^2+x+1)(x^2-x-1) = 0 which again gives the same quadratics.

5:25 The cube roots of unity :)

x⁴ – x² – 2x – 1 = 0 x⁴ – (x² + 2x + 1) = 0 x⁴ – (x + 1)² = 0 (x²)² – (x + 1)² = 0 (x² + x + 1)(x² – x – 1) = 0 Then solve with quadratic formulae to get x = ½(1 ± √5) and ½(1 ± i√3).

the golden ratio as well as cube roots of unity, this is fantastic stuff

This quartic is a difference of squares and very easily solved. The method of "abusing" the quadratic formula almost always makes the original equation more difficult to solve, but in this case it happens to work out neatly.

Next, solve polynomial with degree 6 with cubic equation 😎

x^3 term: allow me to introduce myself

If I am not wrong, this works only if b^2 = 4*a*n, where n is the constant in c (n+x^4). This way, we can get rid of the square root.

It's 5am and I'm watching a math video lol rip.

Pour raccourcir la rédaction remarquons que cette équation peut s'écrire: x⁴ = (x+1)² ce qui donne de nouvelles équations du second degré: x² = x+1 ou x² = –x–1. Etc. Une question se pose: ne serait-ce pas cette particularité qui permet d'utiliser ta méthode ? J'adore toujours autant tes tutos 👍👍👍

Foarte interesant acest mod de a rezolva această ecuație, nu l-am mai întâlnit până acum. Felicitări. Succes in continuare.

I always try the problems before watching the videos. After watching your last video, I had a clue how this would work, so I factored out x^2-1 from the quartic and quadratic terms and used the quadratic formula with a=x^2-1, b=-2, and c=-1. It works just the same as far as resulting in the 2 same quadratics, x^2-x-1=0 and x^2+x+1=0. I love this new to me use of the quadratic formula and I would not call it ABUSE. Did you invent this? If so, congratulations!

before watching: convert to difference of two squares (x^2)^2 - (x+1)^2 then two quadratic factors [x^2 - x - 1][x^2 + x + 1] = 0 . Real roots are phi, Complex roots (-1+- iRoot3)/2

Nice eq. I suggest this solve x^4 - x^2 - 2x - 1 = 0 x^4 - (x+1)^2 = 0 (x^2-x-1)(x^2+x+1)=0 x(1,2) = (1+-5^0.5)/2 x(3,4) = (-1+-i3^0.5)/2

That means the original equation allows a pretty simple factorization. Either notice: x^2 + 2x + 1 = (x+1)^2, so x^4 = (x+1)^2 or (x^2 + x + 1) (x^2 - x - 1) = x^4 - x^2 - 2 x - 1 which has a simple solution in terms of quadratics. There's 2 real and 2 complex roots.

Using this, we can derive an ULTIMATE Fibonacci sequence defined as: a0 = 1 a1 = 1 a2 = 1 a3 = 1 a(n+4) = a(n+2)+2*a(n+1)+an So, the sequence will be 1, 1, 1, 1, 4, 4, 7, 13, 19, 31, 52, 82... compared to the original Fibonacci sequence of 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 231, 375, 606, 981... where the ratio of a(n+1)/an as n goes to infinity also approaches the Golden Ratio.

This is my 1/0 timeth seeing brilliant on a black pen red pen video💀

Does this work on cubic aswell?

i'm in class 8 or grade 8 or standard 8. so I don't know derivatives, differentials, integrations. but I know complex numbers, some trigonometries, logarithms and other things of grade 9-10. blackpenredpen helped me a lot to understand these. still I don't know any calculus topic. i have invented some formulas and theorems like 3d trigonometry with 2 angles and 12 ratios, product of some infinite sums, relation between all types of means/averages like quintic>quartic>cubic>quadratic>arithmetic>harmonic etc. recently I'm trying to make quintic formula with radicals, because I don't know calculus and some special functions. many people thinks , it's not possible and they already have given the proof. but I think there's still a probability that we can make quintic formula with only radicals. blackpenredpen, can you help me by explaining without calculus that why this is'nt possible? please make a video on it.