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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.

NN2
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  • i think you should check out https://math.stackexchange.com/questions/1393869/conditions-for-distinct-real-roots-of-cubic-polynomials?noredirect=1&lq=1 – Aderinsola Joshua Mar 12 '21 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 '21 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 – Aderinsola Joshua Mar 12 '21 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 '21 at 11:36
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    How did you find the necessary conditions and what makes you think that they might not be sufficient? (Applying Vieta's formulas...) – Vepir Mar 12 '21 at 11:39
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    @Vepir Thank you for the hint. – NN2 Mar 12 '21 at 11:51

3 Answers3

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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

Aderinsola Joshua
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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.

JulianKarlBauer
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NN2
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What about rewritting

$$x^3+px^2+qx+r=0$$ $$=y^6+py^4+qy^2+r=0$$ $$=(ay^3+by^2+cy+d)(ay^3-by^2+cy-d)=0$$ with $$ 1= a^2, p = -(b^2-2ac), q=c^2-2bd, r=-d^2$$ and then testing that $$-27a^2d^2 + 18 abcd - 4 ac^3 - 4 b^3 d + b^2 c^2 \ge 0 .$$

The idea being that if and only if a root $x_0$ is positive we can rewrite the factor $(x-x_0)$ as $(\sqrt x+\sqrt x_0)(\sqrt x-\sqrt x_0)$ and then we can look for the existence of three real square roots. It is a travel down to madness, but perhaps it could be useful in some circumstance or for generalisations.

Also, via $-2ac=p+b^2$ and $2bd=-q+c^2$ the test can be rearranged so that all the auxiliary constants are squares, $$ 54 r +9 pq +13 qb^2-5pc^2-7b^2c^2 \ge0$$ but I am a bit dissapointed because I honestly expected to be able to read the signs of $p,q, r$ from here.

arivero
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    You can see the solution in my answer below. About your approach, the first idea (use $x^2$ instead of $x$) is interesting but later it seems complex as the final eaquation involves both $p, q, r$ and $b,c$. – NN2 Jun 07 '23 at 11:22
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    Yes, it is a road to hell, this approach. Still I wanted to post it because the "factorisation" of an equation is an interesting topic. The idea of just proving alternating signs is better, and also generalises well beyond the cubic. What I expected was to get a final equation where alternating signs also implied the positivity of the discriminant. – arivero Jun 07 '23 at 11:27