Can You Solve A Radical Exponential?

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[@NickCraigWood]

The formula given generates the Companion Pell numbers. These are defined by like this (similar to the Fibonacci numbers) a(0) = a(1) = 2 a(n) = 2*a(n-1) + a(n-2) sqrt(2)+1 is known as the silver ratio and plays the same role as the golden ratio does to the Fibonacci numbers. The formula given is the Binet style formula for the Companion Pell numbers. If you knew this obscure fact then the problem is very easy - I suspect the person who invented the problem did!

[@freddyalvaradamaranon304]

Muchas gracias por compartir tan interesante ecuacion exponencial 😊. Buena y didáctica explicación. 😊❤😊.

[@ytlongbeach]

The key was figuring out 1/(sqrt(2)-1) =sqrt(2)+1, then combine the left side under one term, and it’s solvable. How can one solve a problem like: (Sqrt(2)+1)^x + (sqrt(2)-1.1)^x = 34

[@pjb.1775]

The graph surprisingly looks like a parabola though...

[@yakupbuyankara5903]

X=4

[@phill3986]

👏👍😁☮️✌️☮️😁👍👏

[@sheepcommander_]

YOU CAN DO THAT???

[@ShortsOfSyber]

Sure!

[@UpShir67]

Since (√2+1)^x (√2-1)^x = 1 Let r = (√2+1)^x = 1/(√2-1)^x r + 1/r = 34 r² - 34r + 1 = 0 r = ½(34 ±√(34²-4)) r = 17 ±12√2 r = 3²+√8² ±2*3*√8 r = (3±√8)² r = (√2²+1²±2*√2*1)² r = (√2±1)⁴ r = (√2+1)⁴ = (√2+1)^x x = 4★ r = (√2-1)⁴ = 1/(√2-1)^x x = -4★

[@Jiyoon02]

Like 98% of other highschool-arithmetic-disguised pseudo-thought-provoking quizzes, this one is yet another same lame quadratic. The word 'underwhelmed' isn't nearly enough to describe my disappointment. Honestly, why did you make this video? There are already one too many meaningless, trivial quadratic videos, as yourselves, on KZbin. You could have spent your time creating something novel, at least something that isn't this. Instead, you chose to waste it on something so unremarkable

[@ShortsOfSyber]

check @NickCraigWood's comment

[@Don-Ensley]

x ∊ { -4, 4 }