Germany| A Nice Math Olympiad exponential problem | solve for x.

How to solve a Nice Math Olympiad exponential question involving algebra using the laws of indices and algebraic identity. #matholympiadproblem #exponentialequations #algebraic_identities

[@RyanLewis-Johnson-wq6xs]

(1+Sqrt[5]/2)^12=161+72Sqrt[5]

[@chishimbawanchinga8886]

It was ease to follow. Thanks for sharing

[@JJONLINEMATHSCLASSchannel]

Glad it was helpful!

[@Honeyshaf]

Good

[@JJONLINEMATHSCLASSchannel]

Thanks

[@isobar5857]

Oh, my God ...the patience of a saint and an intellect to match. Wish some of it would rub off on me! Enjoy your videos so much dear lady, you are a breath of fresh air. You deserve so many more subscribers.

[@JJONLINEMATHSCLASSchannel]

You are so kind. Thanks for your words of encouragement and love

[@danielnwachukwu9407]

Great job

[@JJONLINEMATHSCLASSchannel]

Thank you

[@garryfrancois2320]

Good job my teacher.

[@JJONLINEMATHSCLASSchannel]

Thanks

[@johnstanley5692]

Alternative? use sequence for Z(n )= X^n + 1/X^n here Z(1)=5^(1/2). (Z(0)=2) Recursion Z(n+1)=Z(n)Z(1)-Z(n-1)
(Also doubling formula Z(2n) = Z(n)^2). Here need to get Z(3) and then doubling formula [ Z(1), Z(2), Z(3) ]=[ 5^(1/2), 3, 2*5^(1/2)].
Now Z(6)=Z(3)^2-2 =18. Z(12)= Z(6)^2-2 =322. So we have, (X^12+ 1/X^12) = 322. => X^12= 161+72*5^(1/2)

[@devonwilson5776]

Greetings. Thanks always for sharing and for pointing out the pitfalls that would have been encountered if one were to have resorted to employ the usage of Pascal's triangle. That was absolutely what came to mind when the expression was reduced to the 6th power. It was indeed a sweet little problem. Thanks again. Blessings.

[@JJONLINEMATHSCLASSchannel]

You are welcome. Thanks so much for engaging.

[@bhaskarbattodi7442]

Super mam

[@JJONLINEMATHSCLASSchannel]

Thanks

[@Honeyshaf]

❤❤

[@JJONLINEMATHSCLASSchannel]

❤️❤️

[@FredJehomiah]

If it's long, that's mathematics for you, there are no shortcuts...kudos🎉 you are an amazing math solver

[@JJONLINEMATHSCLASSchannel]

Wow, thanks!

[@LifnaCabi]

Fantastico e muita alegria.❤❤.
Traduzir para ingles.

[@JJONLINEMATHSCLASSchannel]

Am glad you liked it

[@marcgriselhubert3915]

Much too long!! Be a = (1+sqrt(5))/2
a^3 = (1+3.sqrt(5) +15+5.sqrt(5))/8 = 2+sqrt(5)
So a^6 = (2+sqrt(5))^2 = 4+4.sqrt(5)+5 = 9+4.sqrt(5)
And so a^12 = (9+4.sqrt(5))^2 = 81+72.sqrt(5)+80
Finally a^12 = 161+72.sqrt(5). That's all!

[@labzioui1]

This is a problem that many have discussed, it is a viral problem of this golden ratio.
We want something new
(Thank you)
PS: I can provide you with the links to videos that covered this viral issue.

[@JJONLINEMATHSCLASSchannel]

Thanks for engaging. This is a new channel with a new audience, many are yet to see this . I particularly, just came across it.

[@syedmdabid7191]

Valde Valde facilis! Hic 64 est. Responsi.

[@JJONLINEMATHSCLASSchannel]

??

[@Extratremendouszeus]

That's my black fair lady😅😅😅

[@JJONLINEMATHSCLASSchannel]

Thanks for always engaging

[@Extratremendouszeus]

@@JJONLINEMATHSCLASSchannel You are welcome dear

[@ugurdinc7768]

(3×)"=9×" ? Why 9x?

[@mohamadfazli5575]

Your teaching method of aljabra is definitely confusing,,,

[@JJONLINEMATHSCLASSchannel]

Wow!

[@lawrencemoore2414]

You confused yourself along the way. I wonder what that will do to your students. Besides, you have not shown that your final answer is correct.

[@JJONLINEMATHSCLASSchannel]

If you didn't understand any part , ask questions. There is no confusion anywhere.

[@chukschukwudi9877]

Great job

[@JJONLINEMATHSCLASSchannel]

❤️❤️