Germany | Can you solve this ? | A Nice Math Olympiad Exponential Problem

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[@faustinus23]

Do long division and get x -1 + (1/x) = 3 immediately. After solving the quadratic, rewrite this equation as (x^2 +x+1)(x^2-4x+1)=0 and find the two additional complex roots.

[@davidbrisbane7206]

This equation has four roots. It would have seven roots, but by its construction, the triple root of x = 0 must be excluded.

[@leetucker9938]

awesome channel

[@learncommunolizer]

Thank you!

[@is7728]

Nice approach but some roots have been cancelled since you directly divide them
So there should be 7 roots instead of 2

[@davidbrisbane7206]

Actually, it only has four roots, as by its construction, the triple root of x = 0 must be excluded, otherwise the lefthand side of the equation would be undefined. Two of them are complex and two of them are real.

[@is7728]

@@davidbrisbane7206 Thanks for clarifying

[@carlosotavioderezende8467]

SuperCool 👏👏👏👏👍😃

[@learncommunolizer]

Thanks you 🙏❤️

[@jyothsnapakanati5366]

Really good work, you made my day 😀

[@learncommunolizer]

Your welcome, Thanks 🙏🙏 u

[@soniamariadasilveira7003]

Obrigada por seus ensinamentos!

[@learncommunolizer]

Thank you very much 🙏🙏🙏

[@learncommunolizer]

Thank you very much

[@user-ei6rd7ei7x]

x = 0 isn't a solution, t = x+1/x, (x^2+1+1/x^2)/(x+1+1/x) = 3, x^2+1/x^2+2 = (x+1/x)^2 = t^2, (t^2-1)/(t+1) = 3, t-1 = 3 and t+1 ≠ 0, t = 4, x+1/x = 4, x^2-4x+1 = 0, x = 2+-sqrt(3).

[@user-bw4pw9fd8s]

Hvala❤

[@learncommunolizer]

Thanks you very much 🙏🙏🙏

[@SidneiMV]

divide numerator and denominator by x⁵
(x² + 1/x² + 1)/(x + 1/x + 1) = 3
x + 1x = u => x² + 1/x² = u² - 2
(u² - 1)/(u + 1) = 3
3(u + 1) = (u + 1)(u - 1)
(u + 1)(u - 4) = 0
u = -1 ∨ u = 4/3
u = -1
x + 1/x = -1
x² + x + 1 = 0 => complex roots
u = 4
x + 1/x = 4
x² - 4x + 1 = 0
x = (4 ± 2√3)/2
*x = 2 ± √3*

[@armacham]

That's what I did, but u = -1 does not give any solutions (not even complex solutions) because that would result in dividing by zero

[@CTJ2619]

curious where are you broadcasting from ?

[@muallimeurdu8145]

I guess india or pakistan

[@inm158]

i