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Show that this equation

$x^5+x-10=0$

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

I don't know how I solve it

enter image description here

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    $\begingroup$ 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
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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 $a_0$.

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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.

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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

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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

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  • $\begingroup$ 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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