Find the Values of k for Which the Equation Has Equal Roots
Solution
Given: $$kx(x-2)+6=0$$
$$kx^2-2kx+6=0$$
Here, $$a=k,\quad b=-2k,\quad c=6$$
For equal roots, $$D=b^2-4ac=0$$
$$(-2k)^2-4(k)(6)=0$$
$$4k^2-24k=0$$
$$4k(k-6)=0$$
$$k=0 \quad \text{or} \quad k=6$$
Since the equation must remain quadratic, $$k\neq0$$.
Answer
The value of k for which the equation has equal roots is: $$\boxed{k=6}$$