Find the Values of k for Which the Equation Has Equal Roots
Solution
Given: $$x^2-4kx+k=0$$
Here, $$a=1,\quad b=-4k,\quad c=k$$
For equal roots, $$D=b^2-4ac=0$$
$$(-4k)^2-4(1)(k)=0$$
$$16k^2-4k=0$$
$$4k(4k-1)=0$$
$$k=0 \quad \text{or} \quad k=\frac{1}{4}$$
Answer
The value(s) of k for which the roots are equal is: $$\boxed{k=0 \text{ or } k=\frac{1}{4}}$$