What is the condition that a cubic equation $x^3+ax^2+bx+c=0$ has exactly three positive real root?

Question

What is the condition that a cubic equation $x^3+ax^2+bx+c=0$ has exactly three positive real root?

If $G^2+4H^3<0$ then it has three real roots (where $G=c-ab/3+2a^2/27$, $H=b-a^2/9$).

Then what would be the required condition?

Answer 1

For cubic polynomial to have 3 positive real Roots , we can split question into two parts (1) Roots are real (2) all roots are positive. $$ $$ Let $f(x)=x^3+ax^2+bx+c$. Find $f'(x)$.
$$f'(x)=3x^2+2ax+b=0\;has\,two\,roots$$ $$\therefore 4a^2-12b\ge 0$$ Let $f'(x)=0$. For. $x=\alpha,\beta$. Now for three roots to be real condition will be $f(\alpha).f(\beta)\le 0$. $$. $$ Also for all three roots to be positive point of local maximum should be positive i.e if $\alpha \lt\beta $ then. $\alpha \gt 0$ use both roots of quadratic $f'(x)$ should be positive. Along with it ensure that $f(0)=c\lt 0$ .

Answer 2

Let $\alpha,$ $\beta$ and $\gamma$ be roots of the equation.

There are two cases:

1. $\alpha\in\mathbb R$,$\beta\notin\mathbb R$ and $\gamma\notin\mathbb R.$

Thus, since $\{a,b,c\}\subset\mathbb R$, we see that $\gamma=\overline{\beta},$ which says that $$(\alpha-\beta)^2(\alpha-\gamma)^2(\beta-\gamma)^2=((\alpha-\beta)(\alpha-\overline{\beta}))^2(\beta-\overline\beta)^2=$$ $$=(\alpha^2-2\alpha Re{\beta}+|\beta|^2)^2\cdot\left(-4(Im{\beta})^2\right)<0.$$ 2. $\{\alpha,\beta,\gamma\}\subset\mathbb R.$

Thus, $$(\alpha-\beta)^2(\alpha-\gamma)^2(\beta-\gamma)^2\geq0.$$ Now, let $\alpha+\beta+\gamma=3u=-a,$ $\alpha\beta+\alpha\gamma+\beta\gamma=3v^2=b$ ans $\alpha\beta\gamma=-c=w^3.$

Thus, $$(\alpha-\beta)^2(\alpha-\gamma)^2(\beta-\gamma)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=$$ $$=27\left(3\left(-\frac{a}{3}\right)^2\left(\frac{b}{3}\right)^2-4\left(\frac{b}{3}\right)^3-4\left(-\frac{a}{3}\right)^3(-c)^3+6\left(-\frac{a}{3}\right)\left(\frac{b}{3}\right)(-c)-(-c)^2\right)=$$ $$=a^2b^2-4b^3-4a^3c+18abc-27c^2$$ and we got the following: $$a^2b^2-4b^3-4a^3c+18abc-27c^2\geq0.$$ For positive roots we need also: $$-a>0,$$ $$b>0$$ and $$-c>0.$$ Indeed, since $\alpha\beta\gamma=-c>0,$ we have two cases:

1. $\alpha$, $\beta$ and $\gamma$ are positives.

In this case we got all that we want;

2. Two roots are negative.

Let $\beta<0$ and $\gamma<0$.

Thus, since $$\alpha>-\beta-\gamma,$$ we obtain:$$0<\alpha\beta+\alpha\gamma+\beta\gamma=\alpha(\beta+\gamma)+\beta\gamma<$$ $$<-(\beta+\gamma)^2+\beta\gamma=-(\beta^2-\beta\gamma+\gamma^2)<0,$$ which says that this case is impossible.

Answer 3

Using as CAS and WLOG $a=1$, the Sturm sequence evaluated at $0$ and $\infty$ gives the signs

$$d,1,\\c,1\\9d-bc,3c-b^2\\-b^2 c^2 + 4 c^3 + 4 b^3 d - 18 b c d + 27 d^2$$

(some positive factors omitted; the last term is for a constant polynomial).

There are three real roots in $[0,\infty)$ if the difference of the number of changes of sign at $0$ and at $\infty$ is three.

The condition seems particularly heavy...

Answer 4

Attempt:

$f(x)=x^3+ax^2+bx+c=(x-u)(x-v)(x-w)$ with

$u,v,w >0$;

Let $u<v<w.$

Then $f(x)>0$ for $x >w$; $f(x) <0$ for $x< u$. (Asymptotic behaviour of $f$)

1) $c=-uvw <0$

2) Need a maximum in $(u,v)$, a minimum in $(v,w)$ (Why?).

$f'(x)=3x^2+2ax +b=0$;

$x_{1,2}=\dfrac{-2a\pm\sqrt{4a^2-12b}}{6}$;

where $a^2-3b>0$, and $-2a >0$, since

$u<x_1<v<x_2<w$;

3) Check: $f''(x_1) <0;$

$f''(x_1)=6(x_1)+2a=$

$-\sqrt{4a^2-12b} <0$√

4) Check: $f''(x_2)=$

$6(x_2)+2a=\sqrt{4a^2-12b} >0$√

5) Finally:

$c<0$; $a <0$; $a^2-3b>0$