Equation $x^5+x-10=0$ , as $x\in R$

Question

Show that the equation $$x^5+x-10=0$$ has one positive root, and this root isn't a rational number.

I don't know how I solve it.

Answer 1

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

Answer 2

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 $a_0$.

Answer 3

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.

Answer 4

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

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