Solving a 'Stanford' University entrance exam

[@nalbrt]

Acredite se quiser, no brasil isso é ez pz, e eu sou um cara que mora numa zona muito rural, estudo em colegio público estadual, e achei isso relativamente fácil, quem dera se entrar na faculdade no Brasil fosse tão fácil assim... Mas amei a explicação, voce explicou muito bem, acho legal em ver indianos muito bons em matemática

[@AsadInternationalAcademy]

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[@AsadInternationalAcademy]

Yes, it's easy. thank you so much!

[@nalbrt]

@@AsadInternationalAcademy thank you so much

[@jidehuyghe4051]

Pas évident ! mais avec ténacité Bravo !!

[@AsadInternationalAcademy]

Watch 110+ most important olympiad questions with solutions by just clicking on the link below kzbin.info/aero/PLybCHBiqtqWP-TrcsG21MXXxlknROi514 Don't forget to share this link with your classmates so that they also have a benefit of it. Please, like and subscribe this channel including pressing bell icon to get the notification of new video!

[@AsadInternationalAcademy]

easy dear

[@ellmg4733]

I didn't got the x^4... Why is x^4 = the problem from the beginn?

[@AsadInternationalAcademy]

in fact this problem is x^12 but I reduced it to x^4 then solved finally

[@AsadInternationalAcademy]

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[@emanuele59b]

Simplify by sqrt(6) and the expression is R^12 where R = 2/(Sqrt(5)+1). One observes that R is the inverse of the golden ratio S. The golden ratio solves the equation S^2 -S - 1 = 0 so we can obtain recursively all the S powers by S^2 = S + 1 and we get S^n = Fib(n) * S + Fib( n-1) where Fib is the Fibonacci number. Now Fib(11) = 89 and Fib(12) = 144 so S^12 = 89 + S * 144. Finally we get R^12 = 1 / S^12 = 1 / ( 89 + S * 144). Putting the expression of S inside we get R^12 = 1/ ( 161 + 72* Sqrt(5) ) and rationalizing we get R^12 = ( 161 - 72* Sqrt(5) ) / (161^2 - 72^2 * 5 ). The denominator is 25921-25920 = 1 so the result is = ( 161 - 72* Sqrt(5) ) .

[@AsadInternationalAcademy]

Wonderful dear, thanks

[@AsadInternationalAcademy]

Watch 110+ most important olympiad questions with solutions by just clicking on the link below kzbin.info/aero/PLybCHBiqtqWP-TrcsG21MXXxlknROi514 Don't forget to share this link with your classmates so that they also have a benefit of it. Please, like and subscribe this channel including pressing bell icon to get the notification of new video!

[@walterwen2975]

A Stanford University entrance exam: [(√24)/(√30 + √6)]¹² =? (√24)/(√30 + √6) = (2√6)/[(√5 + 1)√6)] = 2/[(√5 + 1) = [2(√5 - 1)]/(5 - 1) = (√5 - 1)/2 Let: a = (√5 - 1)/2, a² = [(√5 - 1)²]/2² = (5 - 2√5 + 1)/4 = [2 - (√5 - 1)]/2 = 1 - a a⁴ = (a²)² = (1 - a)² = 1 - 2a + a² = 1 - 2a + (1 - a) = 2 - 3a a⁶ = (a²)(a⁴) = (1 - a)(2 - 3a) = 2 - 5a + 3a² = 2 - 5a + 3(1 - a) = 5 - 8a a¹² = (a⁶)² = (5 - 8a)² = 25 - 80a + 64a² = 25 - 80a + 64(1 - a) = 89 - 144a [(√24)/(√30 + √6)]¹² = a¹² = 89 - 144a = 89 - 144[(√5 - 1)/2] = 161 - 72√5 Final answer: [(√24)/(√30 + √6)]¹² = 161 - 72√5

[@AsadInternationalAcademy]

good