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

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, $\sqrt[a]{b-1}=(b-1)^{1/a}=2>3/2.$
