Show that $x^a+x-b=0$ must have only one positive real root and not exceed the $\sqrt[a]{b-1}$

Question

If we take the equation $$x^3+x-3=0$$ and solve it to find the real roots, we will get only one positive real roots which is $(x=1.213411662)$. If we comparison this with $\sqrt[3]{3-1}=1.259921$, we will find that $x$ is less than $\sqrt[3]{3-1}$.This always happens with any value of $a$ so that $a$ any positive real number and $b$ is a positive real number.

So we can ask the following:

1-prove that $x^a+x-b=0$ must have only one positive real root if $a$ positive real number and $b$ is a positive real number greater than $1$.

2-The value of this roots must be less than $\sqrt[a]{b-1}$

3-What happens when $a$ is a complex value.I mean this thing stays right or not?

Answer 1

If $a > 0$, you can use the Intermediate Value Theorem.

If $a < 0$, let $v$ be the minimum value of $x^a + x$ on $(0,\infty)$. Then $x^a + x - b$ has no positive real roots if $b < v$, one if $b = v$ and two if $b > v$.

If $a$ is complex, $x^a + x - b$ is real only for a discrete set of positive real $x$'s, so there will usually be no positive real roots.

Answer 2

If $f(x)=x^a+x-b$ then $f'(x)=ax^{a-1}+1>0,$ since $a\in\mathbb{R}_+.$ Thus $f$ is strictly increasing and so it must take the value $0$ at most one time. Now, since $f(0)=-b<0$ and $f(b)>0$ it follows from the Intermediate Value Theorem that there exists a root in the interval $(0,b).$ So, $f$ has exactly one root.

Finally, if $f(x)=x^1+x-3=2x-3$ has as root $x=3/2.$ However, $\sqrt[a]{b-1}=(b-1)^{1/a}=2>3/2.$