This question frightened 500K+ examinees!

How do you solve this equation? Use this fanatic trick!

[allanflippin]

Great solution! I'm struggling to figure out how I can apply this technique to more general situations.

[ChristianSquare]

I wrote out a simplified expansion of (1+√13)^6 and because all instances of 1^n=1, it cancels out. It leaves me with 1 + 6√13 + 15*13 + 20*13*√13 + 15*13*13 + 6*13*13*√13 + 13*13*13. I combined the similar terms to get (1280*√13)+4928. I include the denominator which would be 2^6=64 and simplify to 20√13 + 77.

[finmat]

Just doing the calculation faster by hand and it's correct

[wes]

(1+√13)/2 is the positive root of x^2–x–3=0, so x^2=x+3; x^4=x^2+6x+9=7x+12;
Thus,

[PatrickLaenen]

Wow! Pretty cool solution! Does this substitution technique always work or does it happen to be a nice method by coincidence for this particular equation?

[cm]

Not hard to just use the expansion of (x+y)^6 and handle the denominator separately

[anastazijaradisavljevic]

😮 Does this way of solving have a name?

[mcumer]

Ho avuto la stessa idea.. Chiamata x la base, 2x=1+sqrt 13 e quindi (2x-1)^2 =13, da cui si ricava x^2=x+3.. Da questa identità si ricava x^4 e x^6.. Procedimento alternativo: la base della potenza ricorda la formula risolutiva dell'equazione di secondo grado con a=1 e b =-1 e delta =13..quindi 1-4c=13 e c= - 3.. Perciò x^2-x-3=0 e quindi x^2=x+3.. Il seguito è uguale

[MathHub]

Wao. Good technique is used to Solve ❤

[老牛吃嫩草-ry]

Is there a way i can send you an interesting puzzle question? Thanks