How to solve a Nice Math Olympiad exponential question involving algebra using the laws of indices and algebraic identity. #matholympiadproblem #exponentialequations #algebraic_identities
Пікірлер: 37
@chishimbawanchinga888610 күн бұрын
It was ease to follow. Thanks for sharing
@JJONLINEMATHSCLASSchannel10 күн бұрын
Glad it was helpful!
@RyanLewis-Johnson-wq6xsАй бұрын
(1+Sqrt[5]/2)^12=161+72Sqrt[5]
@isobar5857Ай бұрын
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.
@JJONLINEMATHSCLASSchannelАй бұрын
You are so kind. Thanks for your words of encouragement and love
@garryfrancois232018 күн бұрын
Good job my teacher.
@JJONLINEMATHSCLASSchannel17 күн бұрын
Thanks
@danielnwachukwu940727 күн бұрын
Great job
@JJONLINEMATHSCLASSchannel27 күн бұрын
Thank you
@johnstanley5692Ай бұрын
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)
@HoneyshafАй бұрын
Good
@JJONLINEMATHSCLASSchannelАй бұрын
Thanks
@devonwilson5776Ай бұрын
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.
@JJONLINEMATHSCLASSchannelАй бұрын
You are welcome. Thanks so much for engaging.
@bhaskarbattodi7442Ай бұрын
Super mam
@JJONLINEMATHSCLASSchannelАй бұрын
Thanks
@FredJehomiah17 күн бұрын
If it's long, that's mathematics for you, there are no shortcuts...kudos🎉 you are an amazing math solver
@JJONLINEMATHSCLASSchannel15 күн бұрын
Wow, thanks!
@LifnaCabiАй бұрын
Fantastico e muita alegria.❤❤. Traduzir para ingles.
@JJONLINEMATHSCLASSchannelАй бұрын
Am glad you liked it
@marcgriselhubert3915Ай бұрын
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!
@HoneyshafАй бұрын
❤❤
@JJONLINEMATHSCLASSchannelАй бұрын
❤️❤️
@syedmdabid7191Ай бұрын
Valde Valde facilis! Hic 64 est. Responsi.
@JJONLINEMATHSCLASSchannelАй бұрын
??
@ExtratremendouszeusАй бұрын
That's my black fair lady😅😅😅
@JJONLINEMATHSCLASSchannelАй бұрын
Thanks for always engaging
@ExtratremendouszeusАй бұрын
@@JJONLINEMATHSCLASSchannel You are welcome dear
@labzioui1Ай бұрын
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.
@JJONLINEMATHSCLASSchannelАй бұрын
Thanks for engaging. This is a new channel with a new audience, many are yet to see this . I particularly, just came across it.
@ugurdinc7768Ай бұрын
(3×)"=9×" ? Why 9x?
@mohamadfazli5575Ай бұрын
Your teaching method of aljabra is definitely confusing,,,
@JJONLINEMATHSCLASSchannelАй бұрын
Wow!
@lawrencemoore241415 күн бұрын
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.
@JJONLINEMATHSCLASSchannel15 күн бұрын
If you didn't understand any part , ask questions. There is no confusion anywhere.