Find the Values of k for Which the Equation Has Equal Roots
Solution
Given: $$kx(x-3)+9=0$$
$$kx^2-3kx+9=0$$
Here, $$a=k,\quad b=-3k,\quad c=9$$
For equal roots, $$D=b^2-4ac=0$$
$$(-3k)^2-4(k)(9)=0$$
$$9k^2-36k=0$$
$$9k(k-4)=0$$
$$k=0 \quad \text{or} \quad k=4$$
Since the equation must remain quadratic, $$k\neq0$$.
Answer
The value of k for which the equation has equal roots is: $$\boxed{k=4}$$