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Let $$f{(x)} = a x^4 - x^3 + ax - a$$ where $a>0$. We know that it contains $4$ roots. Also, by the Descartes' rule of signs, we know that

  1. There are either $1$ or $3$ positive roots.
  2. There is exactly $1$ negative root.

I observed from plotting the function that it may contain exactly one positive root.

Therefore, can we prove/disprove that the function always contain a pair of conjugate complex roots?

Equivalently, can we prove/disprove that it has exactly $1$ positive root?

tia
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Kay
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1 Answers1

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Let $f(x)=0$.

Thus, $$a=\frac{x^3}{x^4+x-1}.$$ Now, by the Descartes' rule of signs we see that $x^4+x-1$ has an unique positive root $x_0$.

Thus, for $x>x_0$ we have $\frac{x^3}{x^4+x-1}>0$ and for $0<x<x_0$ we have $\frac{x^3}{x^4+x-1}<0$.

Now, $$\left(\frac{x^3}{x^4+x-1}\right)'=-\frac{x^2(x^4-2x+3)}{(x^4+x-1)^2}<0$$ for any $x>x_0$, $$\lim_{x\rightarrow x_0^+}\frac{x^3}{x^4+x-1}=+\infty,$$ $$\lim_{x\rightarrow+\infty}\frac{x^3}{x^4+x-1}=0,$$ and $\frac{x^3}{x^4+x-1}$ is a continuous on $(x_0,+\infty),$ which says that the equation $$a=\frac{x^3}{x^4+x-1}$$ or $$f(x)=0$$ has an unique positive root.

By the same way we see that the equation $f(x)=0$ has an unique negative root.

Michael Rozenberg
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