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