Find the Value of k for Which the Roots Are Real and Equal
Solution
Given: $$4x^2-2(k+1)x+(k+4)=0$$
Here, $$a=4,\quad b=-2(k+1),\quad c=k+4$$
For real and equal roots, $$D=b^2-4ac=0$$
$$[-2(k+1)]^2-4(4)(k+4)=0$$
$$4(k+1)^2-16(k+4)=0$$
$$ (k+1)^2-4(k+4)=0 $$
$$ k^2+2k+1-4k-16=0 $$
$$ k^2-2k-15=0 $$
$$ (k-5)(k+3)=0 $$
$$ k=5 \quad \text{or} \quad k=-3 $$
Answer
The value(s) of k for which the roots are real and equal is: $$ \boxed{k=5 \text{ or } k=-3} $$