Two roots between bounds 
Let $a$ and $b$ satisfy $a \geq b>0, a+b=1$.

*

*Prove that if $m$ and $n$ are positive integers with $m<n$, then $a^{m}-$ $a^{n} \geq b^{m}-b^{n}>0$.

*For each positive integer $n$, consider a quadratic function
$$g_{n}(x)=x^{2}-b^{n} x-a^{n} .$$
Show that $g(x)$ has two roots that are in between $-1$ and $1$ .

Hint: Use derivatives.

Question was given to me by my teacher.
I have successfully proved $(1)$ by considering a function $p(x) = a^x - b^x$ then working with its derivative involving the $\log$ function and provided data to reach the conclusion. (I admit that this idea is not original and is borrowed from the proof of $e^{\pi} > {\pi}^e$)
I couldn't figure out $(2)$. Writing $g(x) = x^2 - bx - a$ then its discriminant is $b^2 + 4a > 0$ by provided data so it does have distinct real roots. Checking the bounds $g(1) = 1 - b - a $ which is $0$ (by provided data) which doesn't lead anywhere and hence any attempt of IVT (or rather, Bolzano's Th.) is not possible.
So nothing conclusive is obtained but even if IVT held would it show that the roots are within $-1$ and $1$ ?
$\star$ Is there a theorem which shows that a polynomial must have two roots in some interval or is there some trivial deduction that I am missing on ?
Any hints/comments on how to proceed will suffice. Sorry if something is incorrect in the above reasoning.
 A: $$ g(x) = x^{2} - b^{n}x - a^{n}. $$
There are always two real roots since $\Delta = b^{2n}+4a^{n}.$
Therefore
$$ x_{1,2} = \frac{b^{n} \pm\sqrt\Delta}{2} = \frac{b^{n}}{2}(1\pm\sqrt{1+4(1-b)^{n}b^{-2n}}). $$
Now, the conditions you gave force $b$ to be in the interval $[0;1]$.
Notice that plugging in the boundary values for $b$ will result in both of the roots being between $-1$ and $1$. So now you need to calculate the derivative of the functions representing roots (it suffices to consider only one function with a constant in front of the square root, then only consider what happens when the constant is $-1$ and what happens if it is $1$). Once that is found, you will be able to find a maximum value. I would assume judging by your post that you will now know how to proceed further. Is that right? If not, I will continue the explanation.
A: Simple manipulation of the graph of the quadratic will suffice.
Since the leading coefficient of $g_n$ is $1$,
and the discriminant $b^{2n} + 4a^n$ is always positive,
it suffices to prove that:

*

*$g_n(-1) \ge 0$;


*$g_n(1) \ge 0$;


*$g_n$ attains its minimum somewhere in $(-1, 1)$.
We can finish our proof from these conditions.  Since $g_n$ has a positive leading coefficient and discriminant, the minimum $g_n(b^n/2)$ is negative (or else $g_n$ will be either always positive or only attains $0$ at one point).  Since $g_n(-1) > 0$ and $g_n(1) \ge 0$ but $g_n(b^n/2) < 0$ where $b^n/2 \in (-1, 1)$, $g_n$ has two roots $\alpha \in (-1, b^n/2)$ and $\beta \in (b^n/2, 1]$ by the intermediate value theorem.

The three conditions can be proved as follows.  It can be shown that $\frac12 \le a < 1$ and $0 < b \le \frac12$.
For any $n \in \mathbb{Z}^+$,
\begin{gather*}
g_n(-1) = 1 + b^n - a^n > 1 + 0 - 1 = 0, \\
g_n(1) = 1 - b^n - a^n \ge 1 - (a + b)^n = 0, \tag{*} \label{A}
\end{gather*}
where $\eqref{A}$ can be shown by expanding $(a + b)^n$ using the binomial theorem.
The minimum of $g_n$ is attained at $b^n / 2 \in (0, 1/2] \subset (-1, 1)$.
A: 
For each positive integer n, consider a quadratic function
$g_n(x)=x^2−b^nx−a^n.$
Show that g(x) has two roots that are in between −1 and 1 .

$g_n(1)=1-(a^n+b^n) \implies g_n(1) \geq 0   ( a^n+b^n < (a+b)^n\implies a^n+b^n <1)$
Also, $g_n(-1)=1+b^n-a^n \implies  g_n(2) > 0$ $ [(b^n-a^n)_{min}=-1$ which is possible  only if $a=1$ and $b=0$ but since $b>0$ ,thus $(b^n-a^n) > -1$ ].
Lastly,speaking of $g_n(0)=-a^n \implies g_n(0)<0$.
Thus,the (sample)graph should look something like this:

