Find the Value of k for Which the Roots Are Real and Equal
Solution
Given: $$x^2-2kx+7k-12=0$$
Here, $$a=1,\quad b=-2k,\quad c=7k-12$$
For real and equal roots, $$D=b^2-4ac=0$$
$$(-2k)^2-4(1)(7k-12)=0$$
$$4k^2-28k+48=0$$
$$k^2-7k+12=0$$
$$(k-3)(k-4)=0$$
$$k=3 \quad \text{or} \quad k=4$$
Answer
The value(s) of k for which the roots are real and equal is: $$\boxed{k=3 \text{ or } k=4}$$