Question

In each of the following find the value of k for which the given value is a soluton of the given equation
$\left(i\right)7x^2+kx-3=0,x=\frac{2}{3}\left(ii\right)x^2-x(a+b)+k=0,x=a\left(iii\right)k{x}^2+\sqrt{2x}-4=0,x=\sqrt{2}\left(iv\right)x^{2}3ax+k=0,x=-a$

Answer 1

$\left(i\right)k=-\frac{1}{6}$

putting $x=\frac{2}{3}$ in the equation we get

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

$\frac{28}{9}+k\times \frac{2}{3}=3$

$k\times \frac{2}{3}=\frac{-1}{9}$

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

(ii)k = ab

putting $x=a$ in the equation

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

$k=a^2-a^2+ab=ab$

(iii) k = 1

putting $x=\sqrt{2}$ in the equation we get

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

$2k=4-\sqrt{2\times \sqrt{2}}$

(iv) k = 2a^2\)