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