The equation $(k+4)x^2+(k+1)x+1=0$ has equal roots for which values of $k$?

Equal roots need the discriminant to vanish: $(k+1{)}^2-4(k+4)=k^2-2k-15=0$, i.e. $(k-5)(k+3)=0$, so $k=5$ or $k=-3$. Check $k=5:9x^2+6x+1=(3x+1{)}^2$. Check $k=-3:x^2-2x+1=(x-1{)}^2$. (Flipping both signs of the factors gives the wrong pair −5 and 3.)