# Range of root of cubic equation having least absolute value

Suppose $a$ and $b$ are two positive real numbers such that the roots of the cubic equations $x^3-ax+b=0$ are all real. Let $\alpha$ is a root of this cubic equation with minimum absolute value. Then range of values of $\alpha$ is given by(choose the correct option)

$(A)$ $\frac{2b}{3a}<\alpha \leq \frac{b}{a}$

$(B)$ $\frac{b}{a}<\alpha \leq \frac{3b}{2a}$

$(C)$ $-\frac{b}{a}<\alpha \leq \frac{b}{a}$

$(D)$ $-\frac{b}{a}<\alpha \leq \frac{2b}{3a}$

This question has been asked on this site but answers use Viete's theorem which I am not familiar with.

Now as product of roots is negative, hence we will have either three roots negative or two roots positive and one negative but as sum of roots is $0$, hence we wil have two roots positive and one root negative.

Then I tried finding points of $f'(x)=0$ for $f(x)=x^3-ax+b$ to get $x=\pm \frac{\sqrt{a}}{\sqrt{3}}$ out of which one gives local maxima and other gives local minima. I drew the rough graph and I think the first positive root will be $\alpha$ but how to get desired range?

• FYI Vieta's theorem is basically what you are doing with the products of the roots and sum of roots. A little bit more, but not too difficult to learn Nov 25, 2017 at 18:58
• $\alpha$ should be less than $\sqrt{\frac{a}{3}}$ Nov 26, 2017 at 3:05
• @Ananya Yes looks to me too but question wants us to find range of $\alpha$ Nov 26, 2017 at 3:06
• @Mathematics As roots of $f'(x)=0$ are $\pm u$ where $u=\sqrt{\frac{a}{3}}$, then for three real roots to exist $f(-u)>0$ and $f(u)<0$ Nov 26, 2017 at 4:53

Let $\alpha,\beta,\gamma$ be the solutions of $x^3-ax+b=0$ with $|\alpha|\le|\beta|$ and $|\alpha|\le|\gamma|$.

By Vieta's formula, $$\alpha+\beta+\gamma=0,\quad \alpha\beta+\beta\gamma+\gamma\alpha=-a\lt 0,\quad \alpha\beta\gamma=-b\lt 0$$

Since $\alpha\beta\gamma\lt 0$, we have either that only one of them is negative or that all are negative.

It is impossible that all three are negative because of $\alpha\beta+\beta\gamma+\gamma\alpha\lt 0$.

So, we see that only one of them is negative.

Suppose here that $\alpha\lt 0\lt\beta\le\gamma$ with $|\alpha|\le|\beta|$.

Then, since $\gamma=-\alpha-\beta$, we have $$\alpha\lt 0\lt\beta\le -\frac{\alpha}{2}\implies |\alpha|\le|\beta|\le \left|\frac{\alpha}{2}\right|$$which is impossible.

So, we may suppose that $\beta\lt 0\lt\alpha\le \gamma$ with $|\alpha|\le|\beta|$.

Since $\gamma=-\alpha-\beta$, we have $$\beta\lt 0\lt\alpha\le -\frac{\beta}{2}\tag1$$

Now, since we have $$a=\alpha^2+\alpha\beta+\beta^2\quad\text{and}\quad b=\alpha\beta(\alpha+\beta)$$ and $$(1)\implies \beta\lt\alpha\le-\frac{\beta}{2}\implies (\alpha-\beta)(2\alpha+\beta)\le 0\implies 2\alpha^2\le \alpha\beta+\beta^2$$ we have $$\alpha=\alpha\times 1\gt \alpha\times \frac{(\alpha\beta+\beta^2)}{\alpha^2+(\alpha\beta+\beta^2)}=\frac{\alpha\beta(\alpha+\beta)}{\alpha^2+\alpha\beta+\beta^2}=\frac ba$$

$$\alpha=\alpha\times 1\le \alpha\times\frac{\alpha\beta+\beta^2+(2\alpha\beta+2\beta^2)}{2\alpha^2+(2\alpha\beta+2\beta^2)}=\frac{3\alpha\beta(\alpha+\beta)}{2(\alpha^2+\alpha\beta+\beta^2)}=\frac{3b}{2a}$$

$$\alpha=\alpha\times 1\gt \alpha\times\frac{(2\alpha\beta+2\beta^2)}{3\alpha^2+\alpha\beta+\beta^2+(2\alpha\beta+2\beta^2)}=\frac{2\alpha\beta(\alpha+\beta)}{3(\alpha^2+\alpha\beta+\beta^2)}=\frac{2b}{3a}$$

It follows from these that $\color{red}{(B)}$ is the only correct option.

• Excellent solution
– DXT
Jan 28, 2019 at 5:56