For what value of k, $x^{2} + 2(k-1)x + k+5$ has at least one positive root? For what value of k,  $x^{2} + 2(k-1)x + k+5$ has at least one positive root?
Approach: Case I : Only $1$ positive root, this implies $0$ lies between the roots, so $$f(0)<0$$ and $$D > 0$$
Case II: Both roots positive. It implies $0$ lies behind both the roots. So, $$f(0)>0$$
$$D≥0$$
Also, abscissa of vertex $> 0  $
I did the calculation and found the intersection but its not correct. Please help. Thanks.

You only care about the larger of the two roots - the sign of the smaller root is irrelevant.  So apply the quadratic formula to get the larger root only, which is 
$\frac{-2(k-1)+\sqrt{4(k-1)^2-4(k+5)}}{2} = -k+1+\sqrt{k^2-3k-4}$.  You need the part inside the square root to be $\geq 0$, so $k$ must be $\geq 4$ or $\leq -1$.  Now, if $k\geq 4$, then to have $-k+1+\sqrt{k^2-3k-4}>0$, you require $k^2-2k-4> (k-1)^2$, which is a contradiction.  Alternately, if $k\leq -1$, then $-k+1+\sqrt{k^2-3k-4}$ must be positive, as required.  
So you get the required result whenever $k\leq -1$.
