In each of the following, find the value of k for which the given value is a solution of the given equation:

In each of the following cases find k.

(i) We are given here that,

$7 x^{2}+k x-3=0, x=\frac{2}{3}$

Now, as we know that $x=\frac{2}{3}$ is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting $x=\frac{2}{3}$ in the above equation gives us,

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

$\frac{28+6 k-27}{3}=0$

$6 k=-1$

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

Hence, the value of $k=-\frac{1}{6}$.

(ii) We are given here that,

$x^{2}-x(a+b)+k=0, x=a$

Now, as we know that  is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting  in the above equation gives us,

$a^{2}-a(a+b)+k=0$

$a^{2}-a^{2}-a b+k=0$

$k=a b$

Hence, the value of $k=a b$.

(iii) We are given here that,

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

Now, as we know that $x=\sqrt{2}$ is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting $x=\sqrt{2}$ in the above equation gives us,

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

$2 k+2-4=0$

$2 k=2$

$k=1$

Hence, the value of $k=1$.

(iv) We are given here that,

$x^{2}+3 a x+k=0, x=-a$

Now, as we know that $x=-a$ is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting $x=-a$ in the above equation gives us,

$(-a)^{2}+3 a(-a)+k=0$

$a^{2}-3 a^{2}+k=0$

$k=2 a^{2}$

Hence the value of $k=2 a^{2}$.