# Necessary and sufficient conditions that a cubic equation has three positive real roots

Given the cubic equation $$x^3+px^2+qx+r=0$$

What are the necessary and sufficient conditions that this equation has three positive real roots?

My attempt:

From this answer, the necessary and sufficient conditions that a cubic equation has three real roots is $$-27r^2 + 18 pqr - 4 q^3 - 4 p^3 r + p^2 q^2 \ge 0 \tag{1}$$

In order to make the roots positive, the necessary conditions are $$p <0\tag{2}$$ $$q >0\tag{3}$$ $$r <0\tag{4}$$ but are these conditions sufficient?

PS: Finally, I found the answer.

• i think you should check out math.stackexchange.com/questions/1393869/… Mar 12, 2021 at 10:44
• @AderinsolaJoshua the link you provided is what I gave already in the question. In fact, my question is the condition for having "3 positive roots", not "3 roots".
– NN2
Mar 12, 2021 at 10:57
• a cubic equation will have 3 roots, can be 3 real or 1 real and 2 complex... do you mean 3 Distinct positive root, ohh nice let me check a paper Mar 12, 2021 at 11:28
• @AderinsolaJoshua I meant "3 positive real roots", not necessary distinct roots( In fact, the necessary and sufficient condition for 3 real roots is already on $(1)$ in my question. And if we want to find 3 distinct roots, it suffices to change $\ge$ by $>$ on $(1)$.) I think $(2),(3)(4)$ are sufficient conditions.
– NN2
Mar 12, 2021 at 11:36
• How did you find the necessary conditions and what makes you think that they might not be sufficient? (Applying Vieta's formulas...) Mar 12, 2021 at 11:39

if a cubic should have 3 positive real root, lets say the roots are $$a$$, $$b$$ and $$c$$ then the cubic equation can be written as $$(x-a)\cdot(x-b)\cdot(x-c)$$, now if you expand it $$x^3-(a+b+c)\cdot x^2+(a.b+a.c+b.c)\cdot x-a.b.c$$ if truly $$a ,b, c$$ are positive real, then coeffient of $$x^2$$ is $$< 0$$, coefficient of $$x$$ is $$> 0$$ and coefficient of $$x^0$$ is $$< 0$$ as you have clearly stated, yes it's the sufficient conidition

– NN2
Mar 15, 2021 at 2:09

Let $$a,b,c$$ be three roots of the cubic equations. We will prove that if $$(2),(3)$$ and $$(4)$$ hold, $$a,b,c$$ must be all positive.

Suppose the contradiction, from $$(4)$$ we can suppose that $$a<0,b<0$$ and $$c>0$$.

From $$(2)$$, we have $$c> -a-b$$.

From $$(3)$$, we have $$ab + c(a+b)>0 \implies ab >c(-a-b)>(-a-b)^2 = a^2 +2ab+b^2$$ or $$a^2+ab+b^2 <0$$ (contradiction).

So, $$a,b,c$$ must be all positive.

Then, $$(1),(2),(3)$$ and $$(4)$$ are necessary and sufficient conditions for having all 3 positive roots.