Question

In the following equation, find the value of k for which the given value is a solution of the given equation, the value of k is :
$k{x}^2+\sqrt{2x}-4=0,x=\sqrt{2}$

Answer 1

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

Substituting $x=\sqrt{2}$,

$\Rightarrow$ $k(\sqrt{2}{)}^2+\sqrt{2}\sqrt{2}-4=0$

$\Rightarrow$ $2k+2-4=0$

$\Rightarrow$ $2k-2=0$

$\Rightarrow$ $2k=2$

$\therefore$ $k=1$