Given $ x^5 + x = 10 $ show that A. It has only one root. B. The root lies between 1 and 2 and C. The root is irrational

Question

So the first two are easy. The third one is the one that's tricky. So I tried assuming that the root is rational and let it be $p/q$ but I couldn't go any further or make real progress. More over it's a undergrad entrance question so it won't use really hard math.

Answer 1

C. We want to show that the solution to $$x^5+x=10$$ is irrational. To do that, we can let $f(x)=x^5+x-10$. Since the constant is $-10$ and the leading coefficient is $1$, we know the only potential rational roots are $\pm1, \pm2, \pm5$, or $\pm10$ (i.e. a rational number in the form $\pm\frac{p}{q}$, where $p|10$ and $q|1$). Since the value of the function is non-zero for each of these potential rational roots, we conclude that the root must be irrational.

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For a complete solution to the three questions:

A. We want to show that $$x^5+x=10$$ has only one real solution. To do that, we can let $f(x)=x^5+x-10$. We know that odd degree polynomial functions will have at least one real root. Observe that $f'(x)>0\ \forall x\in\Bbb R\implies f(x)$ is strictly increasing. This shows that $f(x)$ has only one real solution.

B. We want to show that $$x^5+x=10$$ has a solution on the interval $[1,2]$. To do that, we can let $f(x)=x^5+x-10$. Observe that $f(1)=-8$ and $f(2)=24$. By the intermediate value theorem, since $f(x)$ is continuous on the interval $[1,2]$, we have some $c\in[1,2]$ that satisfies $f(1)<f(c)<f(2)$. In other words, there must exist some $c\in[1,2]$ such that $f(c)=0$, since $-8<0<24$.