Olympiad Mathematics /Indians 

It's very simple for the Olympiade

2√x=x 4x=x² x²-4x=0 X=0 or x=4

Nice.

It is the very simple question not asked in Olympiad exam.

There is something special with this equation!!

indians mathematicians know the solution on their birth!

Makes sense. You don't need to square the left side because the terms are identical.

Simply √x + √x = x 2√x = x x² = 4x x² - 4x = 0 x(x - 4) = 0 x = 0 or x = 4 That's it!

И как всегда. Спасибо за просмотр.

Молодец. Пишу комментарий для поддержки.

sqrt(x)+sqrt(x) = x => 2*sqrt(x) = x => 2*sqrt(x) - x = 0 => sqrt(x)[2 - sqrt(x)] = 0 => sqrt(x) = 0, sqrt(x) = 2 => x = 0, 4

Xis 4

x = 4 I teach math and I would never explain it such a convoluted method. Jeez

In India 8 standard student can solve this type of questions 😂😂

It's impossible that it is from Olympiad

Не было сказано про то, что корень должен равняться числу ≥0, а значит x/2 ≥0, и только после этого возводить в квадрат, а так ноль баллов нахуй

x=0 , x=4

X=0 and X=4

Plus two not a root.

x=4

√x+√x=x √x(1+1)=x √x(2)=x √x=x/2 And then we just substitute 4, because this is optimal.

🤣

30 seconds, no pencil...

OBS. x^(1/2)+x^(1/2)=x x+x+2x=x^2 4x=x^2 x=0 or x=4.

X=4

Olympiad?...

Is this a joke or .... a joke ? 🤣🤣🤣

India should leave this job

The first comment has the shortest way of getting x=0 or 4. Dr. Ajit Thakur (USA).