Find the Value of k for Which the Roots Are Real and Equal
Solution
Given: $$4x^2-2(k+1)x+(k+1)=0$$
Here, $$a=4,\quad b=-2(k+1),\quad c=k+1$$
For real and equal roots, $$D=b^2-4ac=0$$
$$[-2(k+1)]^2-4(4)(k+1)=0$$
$$4(k+1)^2-16(k+1)=0$$
$$ (k+1)\big[(k+1)-4\big]=0 $$
$$ (k+1)(k-3)=0 $$
$$ k=-1 \quad \text{or} \quad k=3 $$
Answer
The value(s) of k for which the roots are real and equal is: $$ \boxed{k=-1 \text{ or } k=3} $$