Find the Values of k for Which the Equation Has Equal Roots
Solution
Given: $$kx(x-2\sqrt5)+10=0$$
$$kx^2-2\sqrt5\,kx+10=0$$
Here, $$a=k,\quad b=-2\sqrt5\,k,\quad c=10$$
For equal roots, $$D=b^2-4ac=0$$
$$(-2\sqrt5\,k)^2-4(k)(10)=0$$
$$20k^2-40k=0$$
$$20k(k-2)=0$$
$$k=0 \quad \text{or} \quad k=2$$
Since the equation must remain quadratic, $$k\neq0$$.
Answer
The value of k for which the equation has equal roots is: $$\boxed{k=2}$$