Values of $p$ for which quadratic possess at least one positive root. 
For what values of $p$ would the equation $x^2+2(p-1)x+(p+5)=0,\ \ \{x,p\}\in \mathbb{R}$ possess at least one positive root ?

I tried $$[2(p-1)]^{2}-4(p+5)\geq 0\\~\\
\implies p\geq 4 \cup p\leq -1$$ 
I am not sure how to move on as the discriminant only ensures that the roots are real.
I look for a short and simple way.
I have studied maths up to $12$th grade.
 A: Hint: Consider any $p$ for which the roots are real.  The quadratic will have a positive and negative root if the constant term, $p+5$, is negative. If the constant term is positive, the two roots have the same sign, and we can use the fact that their sum is $-2(p-1)$.

Full solution: 
By my first statement: if $p < -5$, then the roots have opposite sign, so we have at least one positive root.
If $p = -5$, then we have the quadratic $x^2 - 12x$, which has a positive root.
If $-5 < p \leq -1$, then we have two roots of the same sign.  Since the sum of the roots is $-2(p-1) > 0$, we may conclude that both roots are positive.
If $p \geq 4$, then both roots have the same sign.  Since the sum of the roots is $-2(p-1) < 0$, we may conclude that both roots are negative.
So, we have at least one positive root exactly when $p \leq -1$.

Quicker solution:
If $p \leq -1$, the sum of the roots is $-2(p - 1) > 0$. Conclude that we have a positive root.
If $p \geq 4$, then both roots have the same sign since $p+5 > 0$.  Since the sum of the roots is $-2(p-1) < 0$, we may conclude that both roots are negative.
A: Note that the quadratic formula gives that the roots are $$x_1 =\frac{2(1-p) +\sqrt{[2(p−1)]^2−4(p+5)}}{2},\qquad x_2  =\frac{2(1-p) -\sqrt{[2(p−1)]^2−4(p+5)}}{2}.$$
So the restrictions you found on the value of $p$ to ensure that the roots are real can be adjoined to the restrictions required to make at least one of these roots positive. 
For example, in the case that $p\geq 4$,
$2(1-p)<0$, so the second root can never be positive. For the first root to be positive in this case, we need$$2(1-p) +\sqrt{[2(p−1)]^2−4(p+5)} >0,$$
or $$4(p−1)^2−4(p+5) > 4(1-p)^2,$$ 
which is impossible. Thus, there is no case with positive roots if $p\geq 4$.
I leave the other case to you. Let me know if you have any questions.
A: First of all, we need to have
$$(2(p-1))^2-4\cdot 1\cdot (p+5)\ge 0\tag1$$
Under the condition $(1)$, we need that the larger root is positive : 
$$-(p-1)+\sqrt{p^2-3p-4}\gt 0\tag2$$
Now note that $(1)$ and $(2)$ is sufficient.
