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

Answer 1

Compute the required value.

Since, it is given that $x=\sqrt{2}$ is the solution of the quadratic equation $k{x}^2+\sqrt{2}x-4=0$.

Therefore, $x=\sqrt{2}$ satisfy the given quadratic equation.

$$\begin{gathered} k{\left(\sqrt{2}\right)}^2+\sqrt{2}\left(\sqrt{2}\right)-4=0 \\ \Rightarrow 2k+2-4=0 \\ \Rightarrow 2k-2=0 \\ \Rightarrow 2k=2 \\ \Rightarrow k=1 \end{gathered}$$

Hence, the value of $k$ is $1$.