Solving a quintic equation with a non-standard method. An algebraic challenge.

This video is about a quintic equation that can be solved by using algebraic tricks.
Polynomial expressions, algebra, algebraic equations, SyberMath, algebraic manipulations, equations, substitution, Challenging Math Problems, Non-routine Math Problems, Polynomial equations, an algebraic challenge, algebraic identities, an equation,, non-standard methods, quintic equations

[ramaprasadghosh]

writing y for x+1 one gets
(y-2)^5 + (y+2)^5 = 242 y
or
2(y^5 +40y^3 +80y) = 242 y
or
y^5 +40y^3 =41 y
y^2 = 1 being a divisor of this, one gets
y( y^4 - y^2 + 41 y^2-41) =
or y ( y^2 -1)( y^2 +41) = 0
or y = 0, 1, -1 and
complex conjugates as defined by
y^2 +41 = 0, defines the solution space of y.
so x= -1, 0, - 2 defines the real values of x

[davidseed]

If we look for integer solutions, the trying x=0 gives
-1 +243=242
Now in these sorts of problems, ( a pair of high powers)
Then its usual to be able to swap them. So here we make it
-243 +1=-242. When x=-2.
And finally at x=-1. We have
-32+32 = 242(0)
So the three real roots can be found by inspection.

[RexxSchneider]

At 7:00 "When the sum of the coefficients is 1, that means y=1 is a solution." You might want to have another think about that.

[arpansit]

say a = x-1 and b = x+3 then a+b = 2(x+1) and a-b = -4. Now (a^2-b^2)(a^3-b^3) = a^5+b^5 -a^2b^2(a+b).Now RhS can also be written as = 2(x+1)121. a^5+b^5 = (a^2-b^2) (a^3-b^3) + a^2b^2(a+b).
Now take a+b as common after taking u would notice that the RhS is equal to 121(a+b) then in LHs the equation left after taking common will be equal to 121 and u will end up with quadratic equation in ab and then putting (x-1) (x+3) equal to value obtain from quadratic equation. then the common factor end up as a=-b then we have all the solutions.

[holyshit]

I tried to reduce general quartic to biquadratic but i have got sextic which is difficult to solve for me
Sextic which i got is solvable in radicals but i can solve polynomial equations up to quartic

I solved without seeing solution and arrived at exact same solutions yayyyy....

But I didn't even expand (u - 2)^5 and (u - 2)^5....
I used the elementary identity....
a^5 + b^5 = (a + b)*(a^4 - a^3.b + a^2.b^2 - a.b^3 + b^4)
and in this case a + b = 2u, so u = 0 is a ready solution.... Remaining terms are symmetric and lead to bi-quadratic in u^2 + 4....

[Drk]

Nice! I used the same approach before watching your nice video! I loved this problem, i'm learning your tricks from another videos ❤️

[mxsjncv]

I used the formula for n odd, (a^n+b^n) / (a+b) = a^(n-1) - and with some factoring while doing the algebra I got x^4+4x^3+46x^2+84x = 0 which is solvable.

[holyshit]

This quintic can be easily reduced to quartic which can be solved by multiplication two quadratics in general form and comparing coeeficients
or by rewriting quartic as difference of two squares
P(x)=2x(x^2+3x+2)(x^2+2x+42)

[xiaoxiang]

if you let x=2t-1, the calculation will be more simple

[kpt]

sum of coefficients is zero and not 1 mentioned at around 7:15

[comingshoon]

buen video bro.. gracias por el aporte.. saludos desde Chile!

[damiennortier]

"please comment"
OK, I comment

[neuralwarp]

Sorry: 2+ advert breaks means an automatic thumbs down from me