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.

Suppose $x_{1}$ is a real root, then we have that:
$$ (x_{1}+(k-1))^{2} - (k^2-3k-4) = 0 $$
$$(x_{1}+(k-1))^{2} = (k^2-3k-4)$$
$$(k^2-3k-4) \ge 0$$
It's obviously seen that the positive roots are got only when $k \le -1$.
Q.E.D. 
