Question

Solve:

$\sqrt{2}{n}^2-3n-2\sqrt{2}=0$

Answer 1

Given,

$\sqrt{2}{n}^2-3n-2\sqrt{2}=0$

$\sqrt{2}{n}^2-4n+n-2\sqrt{2}=0$

$\sqrt{2}n(n-2\sqrt{2})+1(n-2\sqrt{2})=0$

$(n-2\sqrt{2})(\sqrt{2}n+1)=0$

$\therefore n=2\sqrt{2},-\frac{1}{\sqrt{2}}$