Olympiad Math hexic Equation |Tricks of Solving (x+1)^6=64 | Solving Olympiad Algebra Challenge
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Пікірлер: 13
@mircoceccarelli66894 ай бұрын
👍👍👍
@anderlecht19693 ай бұрын
Nice one. You are a good teacher.
@superacademy247
3 ай бұрын
Thank you! 😃
@sambhavkhandelwal64704 ай бұрын
simple method: (x+1)^6 = 2^6 When the exponent is equal, the bases are equal x+1 = 2 and x+1 = -2 x = 1 and -3
@michaelhuppertz6738
Ай бұрын
This way you only get the real solutions only, you cannot get the complex solutions with this approach.
@mehmethancicek33724 ай бұрын
Thanks
@superacademy247
4 ай бұрын
Welcome
@yuusufliibaan13804 ай бұрын
❤❤❤ thanks 💯💯💯
@superacademy247
4 ай бұрын
Welcome 😊
@dasliebesgluckprinzip96352 ай бұрын
2^6=64 X+1=2 Rest is simpel
@user-kp2rd5qv8g4 ай бұрын
Let z = (x+1)/2. Then, z^6 = 1. So, z = e^(2 pi i n/6) = e^(i pi n/3), n = 0,1,2,3,4,5. So, z = 1, 1/2[ 1 + sqrt(3)i], 1/2[ -1 + sqrt(3)i], -1, - 1/2[ 1 + sqrt(3)i], 1/2[ 1 - sqrt(3)i]. Thus, x = 2z-1 = 1, sqrt(3) i, -2 + sqrt(3)i, -3, -2 - sqrt(3) i, -sqrt(3)i.
Пікірлер: 13
👍👍👍
Nice one. You are a good teacher.
@superacademy247
3 ай бұрын
Thank you! 😃
simple method: (x+1)^6 = 2^6 When the exponent is equal, the bases are equal x+1 = 2 and x+1 = -2 x = 1 and -3
@michaelhuppertz6738
Ай бұрын
This way you only get the real solutions only, you cannot get the complex solutions with this approach.
Thanks
@superacademy247
4 ай бұрын
Welcome
❤❤❤ thanks 💯💯💯
@superacademy247
4 ай бұрын
Welcome 😊
2^6=64 X+1=2 Rest is simpel
Let z = (x+1)/2. Then, z^6 = 1. So, z = e^(2 pi i n/6) = e^(i pi n/3), n = 0,1,2,3,4,5. So, z = 1, 1/2[ 1 + sqrt(3)i], 1/2[ -1 + sqrt(3)i], -1, - 1/2[ 1 + sqrt(3)i], 1/2[ 1 - sqrt(3)i]. Thus, x = 2z-1 = 1, sqrt(3) i, -2 + sqrt(3)i, -3, -2 - sqrt(3) i, -sqrt(3)i.