Show that this equation


Have one positive root and this root isn't rational number

I don't know how I solve it

enter image description here

  • 1
    $\begingroup$ Have you seen the rational root test? $\endgroup$ – user10354138 May 14 at 7:20
  • $\begingroup$ Hi, please be careful in the use of tags. Abstract algebra and especially functional equations has nothing to do with it. Also, any thoughts or intuitions ? $\endgroup$ – Rebellos May 14 at 7:20
  • $\begingroup$ If this equation has a rational root, then this root must be a divisor of $10$ $\endgroup$ – Fareed AF May 14 at 7:23
  • $\begingroup$ And if this equation have a negative root $a$, then $a^5+a$ is for sure also negative; but the equation gives you that $a^5+a=10>0$. So it is impossible for this equation to have a negative root $\endgroup$ – Fareed AF May 14 at 7:33

Hint: let $f(x) = a_nx^n+\cdots+a_1x+a_0$ be a polynomial and $x = \frac{b}{c}\in \mathbb{Q}$ with $\gcd(b,c) = 1$ is a root of $f(x)$, show that $b\mid a_0$ and $c\mid a_n$. Deduce that if $a_n = 1$, then every root in $\mathbb{Q}$ is an integer that divides $a_0$.


As the derivative $4x^2+1$ remains positive, the function is strictly growing and has at most one root.

By the rational root theorem, that root must be a divisor of $10$. But $P(1)<0$ and $P(2)>0$, and no larger divisor can do.

So there is exactly one positive root, which cannot be rational.


By Descartes' rule of signs, there is only one change of sign in the equation: from $+1x$ to $-10.$ Therefore has the equation exactly one positive root.

The absence of rational roots is proven in Hongyi Huang's answer


Calling $f(x)=x^5+x-10$ you have that $f'(x)=5x^4+1>0\,\,,\,\,\forall x$. That implies $f(x)$ is allways increasing. Now, taking $f(1)=-8$ and $f(2)=24$ you can say that your function has one and only real root.

If you supposse x is rational, that is, $x=p/q$ with $p,q\in\mathbb{Z}$ without common divisors, you arrive at

$$\frac{p^5}{q^5}+\frac{p}{q}-10=0\implies p\left(\frac{p^4+q^4}{q^4}\right)=10\,q\in\mathbb{Z}$$

and then $q$ must be a divisor of $p$, which is a contradiction

  • $\begingroup$ Negative constant term is sufficient to prove a positive root. $\endgroup$ – Yves Daoust May 14 at 7:25
  • $\begingroup$ Yes, it's true. I've show that there is one and only real root . $\endgroup$ – popi May 14 at 7:27
  • $\begingroup$ Are you implictly referring to the rational root theorem ? $\endgroup$ – Yves Daoust May 14 at 7:40

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