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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 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!
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.
Your font is very cool 👌
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Math is My ❤
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Автор решает уравн. y³+y²-12=0 ycложнённо: ведь видно, что y=2 и 2 - единственный полож. корень.
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".
Can you explain why you get -8 & -4
Beautiful solution
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Very nice, thank you!
Our pleasure! Thank you to very much
Браво 👍
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4096. Перебором...
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!
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.
х=4096
Sorry 2^12..?
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