Math Olympiad | Wonderful Algebra Problem | VIJAY Maths

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#exponentialproblems #matholympiad #maths #algebra #radical

[@shrikrishnagokhale3557]

Very nice.A good trick.Thanks

[@sumit-mn6ys]

Great explanation

[@prabhushettysangame6601]

Cool solution 🎉

[@superiorlyrics8326]

👏👏👏

[@Rocio62154]

SENSATIONAL!!

[@franciscook5819]

Factorise the initial equation first
S=a²⁰¹⁹-47a²⁰¹¹+a²⁰⁰⁸
=a²⁰⁰⁸(a¹⁶-47a⁸+1) implies we are looking for a⁸
apply (a+1/a)² then square again then again (as in the video)
a⁸+1/a⁸=47 multiply by a⁸
a¹⁶-47a⁸+1=0
so S=a²⁰⁰⁸(0) = 0

[@vijaymaths5483]

Another one beautiful method 👌

[@ezzatabdo5027]

Thanks Professor for your excellent explanation.

[@vijaymaths5483]

Thank you too, for your best 👌 opinion about my educational video

[@padraiggluck2980]

Square both sides of the first equation three times. Then substitute for 47 in the second equation.

[@vijaymaths5483]

Yes 👍

[@john-paulderosa7217]

Solution chasing a problem.

[@vijaymaths5483]

😀

[@RealQinnMalloryu4]

a+a ➖ =a^2 {1+1 ➖/a+a ➖ } ={a^2+2}/a^2=2a^2/a^2 = 2a.(a ➖ 2a+2) (a^2019)^2= a^4361 (47a^2017)^= 16.609a^4189 {a^4361 ➖ 16.6094189}= 16.609a^172 2003+2003 ➖ 4006 {16.609a^172+a^4006}=16.609a4178 4^410^60 3^2a^ 41^1 2^39 2^2^2^22^530^2 3^2a^ 1^12^39^1 1^11^11^15^6^1 3^2a^2^1^1 5^3^2 3^2 2 5^13^2 3^2 2 1^11^1 3^1 2 32 (x ➖ 3x+2)