Find the Values of k for Which the Equation Has Real and Equal Roots
Solution
Given: $$x^2+k(2x+k-1)+2=0$$
$$x^2+2kx+(k^2-k+2)=0$$
Here, $$a=1,\quad b=2k,\quad c=k^2-k+2$$
For real and equal roots, $$D=b^2-4ac=0$$
$$ (2k)^2-4(k^2-k+2)=0 $$
$$ 4k^2-4k^2+4k-8=0 $$
$$ 4k-8=0 $$
$$ k=2 $$
Answer
The value of k for which the roots are real and equal is: $$\boxed{k=2}$$