Math Olympiad Problem | A Nice Math Square Root Simplification Problem | Best Trick.

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How To Solve This Math Olympiad Square Root Simplification Problem,
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[@alaeddinemustapha8496]

When any knowledge goes beyond necessity it loses its importance.

[@varshiniexim3823]

Excellent. Great job ❤

[@jamespn]

Great illustration on methods 1 and 2. Wish I had you as an instructor while in school. God speed on your you tube success.

[@Mathsfocus8610]

Amen. Thanks

[@TheFarmanimalfriend]

well done! 😃

[@Marcos33914]

Very good!

[@HosseinAnsari-wc4ws]

So easy

[@AlCapone-g3d]

Решить уравнение означает найти его корни и доказать, что других корней нет. А разьве у уравнения 31-й степени не должен быть 31 корень включая комплексные?

[@hakade5846]

Is it really Math OLYMPIAD Problem?? REALLY??

[@AlphaSorceror]

They call everything Olympiad problem. These are just indices.

[@에스피-z2g]

I don't think so.

[@AlphaSorceror]

Just count the number of square roots. There are 5 so the LHS index power must be 1 / 2^5) = 1/32
X^(1/32) = 2^3
Then solve.
What nonsense that this is Olympiad?

[@Velvetmystz]

@@AlphaSorcerorisn’t your answer wrong?

[@AlphaSorceror]

@@Velvetmystz it’s wrong. I saw it as 5 sets of square root on an “x” but it isn’t.

[@alaeddinemustapha8496]

Use indes laws.
The LHS = x to the power 1/32.
X = 8 to the 32

[@에스피-z2g]

x^(1/2+1/4+1/8+1/16+1/32)=x^(31/32)
=8
x=8^(32/31)

[@GREGORIOMAXIMIANODONASCIMENTO]

ok

[@DxS-v5n]

x=2^(32/31*3)= 8*(8^1/31)

[@gerhardb1227]

subst u^32 = sqrt(x)
=> u^31= 8
u = 8^(1/31)
resubstitution:
sqrt(x) = 8^(32/31)
x = 8^(64/62) = 8^(32/31)

[@alexniklas8777]

Мало знаков, добавить ещё 5😅😅😅

[@murattanyel1029]

Another way is by moving the x's under the next square root by squaring them. That ends up as x^(31) under five square roots. 2^5 is 32, so we end up with x^(31/32) = 8, just like the second method.

[@mikeymikey8079]

√x √x √x √x √x = 8 Squaring both sides =>
x √x √x √x √x = 64 => (√x √x √x √x) = 64 / x
substitute back into original equation =>
√( x * 64 / x ) = √64 = 8

[@ganeshdas3174]

Apply thumbs rule (1 - 1/2^n) where n = # of square roots (√)
√ = 1/2, √√ = 3/4, √√√ = 7/8...
√√√√√=31/32

[@Ттрр123]

1/32+1/16+1/8+1/4+1/2....

[@prollysine]

x^(31/32)=8 , /()^(32/31) , x=8^(32/31) ,

[@kariunas13]

Last sq.root need
X^1/2

[@subratabiswas2502]

√(x√x........)=x^(31/32) ; so, x^(31/32)=8 ; i,e x=8^(32/31) ; i,e x=2^(3×32/31) ; i,e x= 2^(96/31)

[@bitotzky]

√x√x√x√x√x= 8
√x.x^2√x√x√x=8
√x.x^2.x^4√x√x=8
√x.x^2.x^4.x^8√x=8
√x.x^2.x^4.x^8.x^16=8
√x^(1+2+4+8+16)=8
√x^31=8
x^31/32=8
X= 8^(32/31)
Los índices de cada raíz se van multiplicando 2x2=4, 4x2=8, 8x2=16, 16x2= 32

[@Mr.Grey.39]

Not series

[@HectorLister-y9d]

Esta mal

[@ericasi4562]

Это шутка?

[@birmanmanurung6596]

gak jls prblem mu bos sudalah bermain angka dah gak jls