For what value of k are the roots of the quadratic equation

The given equation is

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

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

This is of the form $a x^{2}+b x+c=0$, where $a=k, b=-2 \sqrt{5} k$ and $c=10$.

$\therefore D=b^{2}-4 a c=(-2 \sqrt{5} k)^{2}-4 \times k \times 10=20 k^{2}-40 k$

The given equation will have real and equal roots if D = 0.

$\therefore 20 k^{2}-40 k=0$

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

$\Rightarrow k=0$ or $k-2=0$

$\Rightarrow k=0$ or $k=2$

But, for k = 0, we get 10 = 0, which is not true.

Hence, 2 is the required value of k.