Question

The value of k for which the roots are real and equal of the following equation
$kx(x-2\sqrt{5})+10=0$ is -----

Answer 1

$kx(x-2\sqrt{5})+10=0$

$\Rightarrow$ $k{x}^2-2\sqrt{5}kx+10=0$

$\Rightarrow$ We have, $a=k,b=-2\sqrt{5}k,c=10$

$\Rightarrow$ It is given that roots are real and equal.

$\therefore$ $b^2-4ac=0$

$\Rightarrow$ $(-2\sqrt{5}k{)}^2-4(k\left)\right(10)=0$

$\Rightarrow$ $20k^2-40k=0$

$\Rightarrow$ $20k(k-2)=0$

$\Rightarrow$ $20k=0$ and $k-2=0$

$\therefore$ $k=0$ and $k=2$