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