Math Olympiad | Wonderful Algebra Problem

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Math Olympiad | Wonderful Algebra Problem | VIJAY Maths

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Ай бұрын

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Пікірлер: 14

@superiorlyrics8326 Ай бұрын

👏👏👏

@shrikrishnagokhale3557 Ай бұрын

Very nice.A good trick.Thanks

@sumit-mn6ys Ай бұрын

Great explanation

@prabhushettysangame6601 Ай бұрын

Cool solution 🎉

@Rocio62154 Ай бұрын

SENSATIONAL!!

@franciscook5819 Ай бұрын

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 Ай бұрын

Another one beautiful method 👌

@ezzatabdo5027 Ай бұрын

Thanks Professor for your excellent explanation.

@vijaymaths5483 Ай бұрын

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

@padraiggluck2980 Ай бұрын

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

@vijaymaths5483 Ай бұрын

Yes 👍

@john-paulderosa7217 Ай бұрын

Solution chasing a problem.

@vijaymaths5483 Ай бұрын

😀

@RealQinnMalloryu4 Ай бұрын

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)