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.

The roots are given by $1-k\pm\sqrt{k^2-3k-4}$, for which we have:
$$\cases{
1 - k + \sqrt{k^2 - 3k - 4} > 0 & if $\phantom{~-5< \;}k \le -1$\\
1 - k - \sqrt{k^2 - 3k - 4} > 0 & if $~-5<k\le -1 $.
}
$$
Wolfram Alpha gives the plus and 
subtract cases. So for $k\le -1$, you get at least one positive root.
