In each of the following, find the value of $k$ for which the given value is a solution of the given equation:
$kx^2+\sqrt{2}x-4=0$, $x=\sqrt2$

Given:

Given equation is $kx^2+\sqrt{2}x-4=0$.

To do:

We have to find the value of $k$ for which $x=\sqrt{2}$ is a solution of $kx^2+\sqrt{2}x-4=0$.

Solution:

If $x=m$ is a solution of $f(x)$ then $f(m)=0$.  

Therefore,

For $x=\sqrt{2}$

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

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

$2k+2-4=0$

$2k-2=0$

$2k=2$

$k=1$

The value of $k$ is $1$.

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Updated on: 10-Oct-2022

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