In each of the following, find the value of $k$ for which the given value is a solution of the given equation:
$7x^2+kx-3=0$, $x=\frac{2}{3}$

Given:

Given equation is $7x^2+kx-3=0$.

To do:

We have to find the value of $k$ for which $x=\frac{2}{3}$ is a solution of $7x^2+kx-3=0$.

Solution:

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

Therefore,

For $x=\frac{2}{3}$

$7x^2+kx-3=0$

$7(\frac{2}{3})^2+k(\frac{2}{3})-3=0$

$7(\frac{4}{9})+k(\frac{2}{3})-3=0$

$9(\frac{28}{9})+9k(\frac{2}{3})-9(3)=9(0)$

$28+6k-27=0$

$6k+1=0$

$6k=-1$

$k=\frac{-1}{6}$

The value of $k$ is $\frac{-1}{6}$.

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Updated on: 10-Oct-2022

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