Solving A Quintic Equation without Using the Quintic Formula

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SIMILAR PROBLEM

#polynomials #polynomialequations
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(1+z)^5=(1-z)^5

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[randomjin]

∀a, b ∈ ℂ, k ∈ ℤ, k ≠ 0, the equation (x+a)ᵏ = (x+b)ᵏ is reducible to k equations of the form y = zₘ where y = (x+a)/(x-b) and zₘ = cis(2𝜋/m) is the k-th root of unity and indexed 0 ≤ m ≤ k-1. Each of those k equations are simple linear equations in x. If k < 0 we take -k as a basis. For this video, a = 1, b =-1, k = 5.

[maitrehenryalain]

There is a mistake in the 2nd method. The numerator of z should be : exp^(2n*pi*i)/5 - 1.
So, at the end, z = i * tan ((2n * pi * i)/5) (no minus sign)

[HenkVanLeeuwen-io]

I like the 1st method better because its simpler having square roots. The 2nd had trigonometric functions.

[vzaimo]

Hello, Syber! Thank you for videos! Can you solve x^(e^x)=e? It's my equation, from Russia.

[pavelgatnar]

Just expand both statements. Odd parts cancel, even parts duplicate.
2z^5+20z^3+10z=0.
z((z²)²+10z²+5)=0 =>z=0
y=z², z=±√y => y²+10y+5=0.

[fabianperdomoborja]

Hello, you are very skillful
Greetings from Colombia

[ajabkhan]

Ajab khan khattak.I solved this question mentally.( 1+z)^5= (1-z)^5.Looking at it, it is clear that z should be =0, meaning 1+0=1-0 so the answer is 0.

[wes]

I used the first method but the second method is clever and easier in the long run for those familiar with the exponential formats.

[lawrencejelsma]

Something is strange with the 2nd method answer. The answer suggests only half of the Euler Circle came up with the answer. I was expecting since tan(x) over 2π is two graphs we would get Zn = 0 and the other points over all four quadrants. Versus the tan() in your solution being as if we are looking at tan()s 0 to π or 1st and 2nd quadrants. I was expecting from 0 and including 0 and complex conjugate pairs of quadrant 1 to quadrant 4 and quadrant 2 to quadrant 3 pairs. Maybe the solution Zn = -i tan(nπ/5) equivalence is telling the same story since it all looks correct with five roots and including the 0 root to consider partitions 1, 2, 3 and 4 complex pairs divisions around the Euler Circle. I'll have to think more in your 2nd method if we can make it a two quadrant because of symmetry tan function solution as your clever solution seems to do representing all four quadrants.

[scottleung]

I think you forgot to mention z=+-i*sqrt(2*sqrt(5)-5) in the first method.

[JosBergervoet]

The title (about "the quintic formula") is strange... Galois showed there doesn't exist a quintic formula!

“Using the Quintic Formula”? I thought there was no such thing?

[deathbyvinyl]

This is obviously derived from a paper, you should at least link it.