A Quintic Equation Solved Without The Quintic Formula

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#ChallengingMathProblems #PolynomialEquations

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[jim]

"Do not try and use the quintic formula, that's impossible. Instead, only try to realise the truth... there is no quintic formula."

[loneranger]

Without Quintic formula got me lol. Earned a like just because of the title

[richardryan]

The title of this video is a little misleading. It has been proved that a formula for solving all quintic equations, which uses a finite number of radicals and basic operations, such as addition and multiplication, does not exist. For example, the solutions to x^5 − x + 1 = 0 cannot be expressed using radicals.

[GillAgainsIsland]

I used binomial expansion for (x+1)^5. Then cross multiply and simplified to get a 5th degree polynomial and set it equal to zero. Then, using the rational roots theorem I quickly found x=2 was one solution. Then used synthetic division to find the 4th degree polynomial, etc. Only took 5 minutes to find all the solutions, with x= -1 rejected since it's extraneous.

[sandorszabo]

It's a quartic equation. In the first step we try to determine the rational solutions.

[vishalmishra]

1. (x+1)^5 / (x^5+1) = (1+1/x)^5 / (1 + 1/x^5). By symmetry, roots will be reciprocal pairs.
2. If x is an integer, then 11 is a multiple of x^5+1. 32+1=33 is a multiple of 11, so x = 2. So, 1/x can also be 2. So, x = 2 or 1/2. We found 2 roots already.
3. (a-b) is a common factor of (a-b)^n / (a^n-b^n). So, (x+1) can be cancelled from (x+1)^5 / (x^5+1) to get (1 4 6 4 1) / (1 -1 1 -1 1). Now Quintic has become Quartic.
4. 11 ( 1 4 6 4 1) = (11 44 66 44 11) and 81 (1 -1 1 -1 1) = (81 -81 81 -81 81). Therefore, (81 -81 81 -81 81) - (11 44 66 44 11) = (70 -125 15 -125 70) . (x^4 x^3 x^2 x 1). Divide coefficients by 5 and terms by x^2.
5. Quartic becomes - 14 (x^2 + 1/x^2) - 25 (x + 1/x) + 3 = 0. Then become Quadratic - 14 (x+1/x)^2 - 25 (x+1/x) - 25 = 0.
6. Solving Quadratic - x + 1/x = [ 25 +/- sqrt(625 + 56x25 = 5^2 . 81 = 45^2)] / 28 = 70/28, -20/28 = 2.5 or -5/7.
7. Solve for x: x^2 + 1 = 2.5x or - 5x/7 => x = 2, 1/2 OR (-5 +/- 3i sqrt(19))/14. Note that last 2 are both complex conjugate and reciprocals of each other (since 25 + 171 = 196 = 14^2) as expected from #1 above.
*Simple* . Right ?

[dominiquebercot]

Bonjour, j’ai une autre idée
X=2 est solution évidente, et en posant x=1/t, on voit que x=1/2 est solution aussi.
Le polynôme vous avez calculé se divise donc par x-2, puis par x-1/2
On trouve70x^2+50x+70=0, qui n’a pas de solution dans R

[timeonly]

When you were solving for u1... knowing that x=2 is a soln above, you could've set u = 2+1/2 = 5/2. Using synthetic div with divisor of u=5/2 on (14, -25, -25), you get 14u+10=0 => u2 = -5/7. Etc...

[jarikosonen]

It looks like the lim (x+1)^5/(x^5+1) = 0, when x -> -1... (not 81/11). The 0/0 looks very uncertain cases. Somehow the 5th root is lost due to the division by zero...

[leopeng]

we can divide x^5 in both numerator and denominator of the left side. so we get ((1/x)+1)^5/((1/x)^5+1)=81/11, which means if x is a solution of the equation, then 1/x must also be a solution. you guess and verify x=2 is a solution, so I know x=1/2 is also a solution. then x-2 and 2x-1are the factors of equation. use long division, we get a standard quadratic equation and then use formula...

[MathElite]

Nice question, quintic formula would be so long if it existed

[albertmcchan]

70x^4 - 125x^3 + 15x^2 - 125x + 70 = 0
We know 2 roots are 2, 1/2
sum of 4 roots: 2 + 1/2 + x + 1/x = 125/70
x + 1/x = -5/7
abs(-5/7) < 2 => other 2 roots are complex.

[michaelempeigne]

no need to divide by x^2 since you know one factor is ( x - 2 ) to reduct it to a cubic.

[musatebi]

As we know +2 is a solution, would not be easier to divide the equation starting with 70x to the fourth power by (x-2) and use the general formula for the remaining equation starting with x cube?

[sidimohamedbenelmalih]

Really really NICE with capital letter

[felipegustavo]

Great video! What software do you use to write that way ("manuscript")? It feels like you are handrighting instead of using a mouse and keyboard, it's very nice!

[-basicmaths]

(x+1)^5/x^5+1=81/11
Divide LHS by(x+1)
(x+1)^4/(x^5+1)/x+1 =81/11
Now we have two equation
(x+1)^4=81
x+1=3, x=2.
We recheck in another equation.
x^5+1/(x+1)=11
32+1)/3=11

[takyc]

I was like - I swear there’s no quintic formula! got me searching google lmao

[ashrafmourad]

wounderful, please Iwant complet explanation about the number theorem

[manojsurya]

At 1:40
Syber math: I'm not gonna skip steps
Also syber math: skips complete factorisation of x⁵+1







Just for fun, its a good explanation