A Nice Math Olympiad Algebra Problem. X=?

Рет қаралды 17,404

Күн бұрын

Пікірлер: 24

@BP-gn2cl Ай бұрын

Step 1: (x^3) ^2=2^2 Step 2: x^3=+/-2 Step 3: x=+/-(2) ^1/3 as x is real. The end Required only 30 seconds. Ninth std math

@mayaq8324 Ай бұрын

when I need to clear my mind, I go here 🙏

@امیدکاظمیشاد Ай бұрын

A Persian proverb says: You turned the morsel around your head

@SALogics Ай бұрын

Nice explanation! ❤❤

@loloflores123 16 күн бұрын

Mate, in 0:45. Just (x³)=√4 ergo x=³√+-2. There, solved.

@gediminasgudavicius1267 Ай бұрын

Thank you for showing how to make 2 into cube, didn't know this.

@vishnuacharya6352 Ай бұрын

Agreed: seems stupidly long-drawn.

@kaustav20062000 Ай бұрын

Make a simple simple things to complex

@SL-up5pl Ай бұрын

I couldn’t agree more.

@is7728 Ай бұрын

Then how do u find all 6 roots (and verify that 4 of them are non-real)

@indianmaid70 Ай бұрын

This is much too easy - it goes way more complicated!!!

@Math_Bee Ай бұрын

I can do it in 3 lines

@pas6295 Ай бұрын

Xis Plus or Minus 2/3.

@LANNGUYEN-kv6jl Ай бұрын

It's a stupid way to solve this equation for x.

@БорисШалагінов Ай бұрын

Если решение получается более громоздким и сложным, чем исходное условие, то это не решение. Настоящее решение - это артистическое и изящное появление фокусника в ослепительно белом костюме и перчатках. А у Вас - нагромождение неубранного мусора.

@Anthony-tp6fm Ай бұрын

Hello ❤❤❤

@learncommunolizer Ай бұрын

Hello 😊 ❤️❤️❤️

@kyintegralson9656 Ай бұрын

Why discard the complex roots? We can solve the 2 quadratic equations, or alternatively realize that in the complex plane, the 6th root of a real number has 6 solutions such that x=(4)^(⅙)·e^(i2πn/6) for n=0,±1,±2,3. n=0,3 give the 2 real solutions. So, the 4 complex roots are x=(±1±i√3)·2^(-⅔)

@BP-gn2cl Ай бұрын

No it is given x is real

@kyintegralson9656 Ай бұрын

@@BP-gn2cl I am going beyond the constraints of the problem; it's educational.

@BP-gn2cl Ай бұрын

@@kyintegralson9656 then its ok