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Very elegantly done such that a retired mathematics professor like myself cam appreciate. Dr. Ajit Thakur (USA).
You should not stop if y is complex. the 12th power of y may not be complex.
Brilliant
Thanks you very much!
Math is My ❤
Thank you very much!
Your font is very cool 👌
Nice video. I'm not sure if people would immediately see that 12=2^3+2^2. And putting 2^10 in there is totally random, people would have to put that in a calculator just as much as they would 2^12 to come to the number of 4096.
Very nice, thank you!
Our pleasure! Thank you to very much
Beautiful solution
Can you explain why you get -8 & -4
Автор решает уравн. y³+y²-12=0 ycложнённо: ведь видно, что y=2 и 2 - единственный полож. корень.
Браво 👍
Thank you very much 👍👍👍
4096. Перебором...
4096.
now solve 4th root of x + 6th root of x = 13.
That's easy, since you specified x = 13.The 4th root of x is ⁴√13.The 6th root of x is ⁶√13.Thus, the sum is ⁴√13 + ⁶√13.
Excelente !
х=4096
These so called Olympiad problems are surprisingly easy
ofc it is easy when you see the answer but when you have to do all of this in a limited time, the real challenge begins.
Sorry 2^12..?
سوال قشنگی بود ولی بهتر نبود جای اون همه کار ریشه درجه سه رو حدس میزدی و با تقسیم حل میکردی
I'll guess that x = 2¹² = 4096 :⁴√2¹² + ⁶√2¹² = 2³ + 2² = 8 + 4 = 12Let's formally find all solutions:⁴√x + ⁶√x = 12¹²√x³ + ¹²√x² = 12This is a cubic equation in y = ¹²√x :y³ + y² = 12Let's do a quick check with factorization:y²*(y + 1) = 2²*3It's clear that y = 2 is a solution. We know (y - 2) is a factor below:y³ + y² - 12 = 0(y - 2)*(y² + 3*y + 6) = 0y = 2, (-3 ± i*√15)/2x = y¹² = 2¹², (-3 ± i*√15)¹²/2¹²Let's manually calculate this as (((y²)*y)²)² :y¹ = 2, (-3 ± i*√15)/2y² = 4, -3*(1 ± i*√15)/2y³ = 8, 3*(9 ± i*√15)/2y⁶ = 64, 27*(11 ± i*3*√15)/2y¹² = 4096, 729*(-7 ± i*33*√15)/2Thus:x = 4096, 729*(-7 ± i*33*√15)/2Let's add y² and y³ from the lines listed above:y² = 4, -3*(1 ± i*√15)/2y³ = 8, 3*(9 ± i*√15)/2y²+y³ = 12, 12 ± 0This checks out.BTW: You said "one hundred twenty four" for 1024, and "four hundred ninety six" for 4096. You should've said "thousand" instead of "hundred".
Very elegantly done such that a retired mathematics professor like myself cam appreciate. Dr. Ajit Thakur (USA).
You should not stop if y is complex. the 12th power of y may not be complex.
Brilliant
Thanks you very much!
Math is My ❤
Thank you very much!
Your font is very cool 👌
Thank you very much!
Nice video. I'm not sure if people would immediately see that 12=2^3+2^2. And putting 2^10 in there is totally random, people would have to put that in a calculator just as much as they would 2^12 to come to the number of 4096.
Very nice, thank you!
Our pleasure! Thank you to very much
Beautiful solution
Thank you very much!
Can you explain why you get -8 & -4
Автор решает уравн. y³+y²-12=0 ycложнённо: ведь видно, что y=2 и 2 - единственный полож. корень.
Браво 👍
Thank you very much 👍👍👍
4096. Перебором...
4096.
now solve 4th root of x + 6th root of x = 13.
That's easy, since you specified x = 13.
The 4th root of x is ⁴√13.
The 6th root of x is ⁶√13.
Thus, the sum is ⁴√13 + ⁶√13.
Excelente !
Thank you very much!
х=4096
These so called Olympiad problems are surprisingly easy
ofc it is easy when you see the answer but when you have to do all of this in a limited time, the real challenge begins.
Sorry 2^12..?
سوال قشنگی بود ولی بهتر نبود جای اون همه کار ریشه درجه سه رو حدس میزدی و با تقسیم حل میکردی
I'll guess that x = 2¹² = 4096 :
⁴√2¹² + ⁶√2¹² = 2³ + 2² = 8 + 4 = 12
Let's formally find all solutions:
⁴√x + ⁶√x = 12
¹²√x³ + ¹²√x² = 12
This is a cubic equation in y = ¹²√x :
y³ + y² = 12
Let's do a quick check with factorization:
y²*(y + 1) = 2²*3
It's clear that y = 2 is a solution. We know (y - 2) is a factor below:
y³ + y² - 12 = 0
(y - 2)*(y² + 3*y + 6) = 0
y = 2, (-3 ± i*√15)/2
x = y¹² = 2¹², (-3 ± i*√15)¹²/2¹²
Let's manually calculate this as (((y²)*y)²)² :
y¹ = 2, (-3 ± i*√15)/2
y² = 4, -3*(1 ± i*√15)/2
y³ = 8, 3*(9 ± i*√15)/2
y⁶ = 64, 27*(11 ± i*3*√15)/2
y¹² = 4096, 729*(-7 ± i*33*√15)/2
Thus:
x = 4096, 729*(-7 ± i*33*√15)/2
Let's add y² and y³ from the lines listed above:
y² = 4, -3*(1 ± i*√15)/2
y³ = 8, 3*(9 ± i*√15)/2
y²+y³ = 12, 12 ± 0
This checks out.
BTW: You said "one hundred twenty four" for 1024, and "four hundred ninety six" for 4096. You should've said "thousand" instead of "hundred".