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.
Best Answer
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.