Thanks to mathlove's comments to the original query which indicated that I had misinterpreted the problem. The answer has been corrected.
I upvoted the original query, for the very good work. However, I prefer the
approach of solving for
$p$ and $q$, given any fixed value of $a$, and then
analyzing further.
$\{p,q\} = \frac{1}{2} \times
[-(a +2) \pm \sqrt{(a+2)^2 - 4(a^2 - a + 2)}]$
$ = \frac{1}{2} \times
[-(a +2) \pm \sqrt{(a^2 + 4a + 4) - 4(a^2 - a + 2)}]$
$ = \frac{1}{2} \times
[-(a +2) \pm \sqrt{(-3)a^2 + 8a - 4)}].$
Since $p,q$ are required to be real, with $p < q$, it is required that
$h(a) = [(-3)a^2 + 8a - 4)]$ must be positive.
That is, if $h(a) = 0$, this would cause $p = q$, which is forbidden by the premise to the problem.
Solving for $h(a) = 0$ gives
$a = \frac{1}{-6} [-8 \pm \sqrt{64 - 48}] $
$= \frac{1}{-6} [-8 \pm \sqrt{16}] $
$= \frac{1}{-3} [-4 \pm 2] $
This means that $a$ must be in the open interval
$(\frac{2}{3}, 2).$
Note : It is possible that different values for $a$ might be used
to maximize the values for $q$ and $p$.
For $a$ in the range, $(\frac{2}{3}, 2).$
The larger root and the smaller root must be separately maximized.
What needs to be maximized are the two values of
$g(a) = [-(a +2) \pm \sqrt{(-3)a^2 + 8a - 4)}]$.
Unfortunately, I was never taught any pre-calculus method for
maximizing the two roots of $g(a)$. In terms of calculus, we have that
$\displaystyle g'(a) = -1 \pm
\left[\frac{1}{2} \times \frac{-6a + 8}{\sqrt{(-3)a^2 + 8a - 4)}}\right].$
I must find all values of $a$ such that $g'(a) = 0.$
In order to do this, I must solve
$(-1) \times \sqrt{(-3)a^2 + 8a - 4} \pm
\left[\frac{1}{2} \times (-6a + 8)\right] = 0.$
This means that $(4) \times [(-3)a^2 + 8a - 4)]$ must = $(-6a + 8)^2.$
This means that $(-12a^2 + 32a - 16)]$ must = $(36a^2 -96a + 64).$
This means that $0 = (48a^2 -128a + 80).$
This means that $0 = (3a^2 -8a + 5).$
This means that $a = \frac{1}{6} \times \left[ 8 \pm \sqrt{64 - 60}\right].$
This means that $a = \frac{1}{6} \times \left[ 8 \pm 2\right].$
This means that the only values for $a$ that might yield $g'(a) = 0$ are
$a = \frac{10}{6}$ or $a = 1.$
Technically, Calculus would require that I now take the 2nd derivative of $g(a).$
However, that is very messy and fortunately, there is a shortcut.
In addition to the possible critical points of
$a = \frac{10}{6}$ or $a = 1$
the boundary points of $a = \frac{2}{3}$ and $a = 2$
can also be considered.
However, when considering the boundary points of $a = \frac{2}{3}$ and $a = 2$,
although $a$ may approach one of the boundary points, it can never equal either of the boundary points.
This means that I have to calculate the two roots, $p$ and $q$ for each of the
4 values of $a$ above. Then, I must select the maximum value for the larger root $q$ and the maximum value for the smaller root $p$, when $a$ is restricted to the open interval $(2/3, 2).$
Setting $a = \frac{2}{3}$ yields $\{q,p\}$ of
$ = \frac{1}{2} \times
\left[-\frac{8}{3} \pm \sqrt{\frac{-12}{9} + \frac{16}{3} - 4}\right]$
$ = \frac{1}{2} \times
\left[-\frac{8}{3} \pm \sqrt{\frac{-12 + 48 - 36}{9}}\right]$
$ = [-\frac{4}{3} \pm 0].$
In fact, this means that as $a$ approaches 2/3, the smaller root will approach (-4/3) from below and the larger root will approach (-4/3) from above.
.....
Setting $a = 1$ yields $\{q,p\}$ of
$ = \frac{1}{2} \times
\left[-3 \pm \sqrt{-3 + 8 - 4}\right]$
$ = \frac{1}{2} \times
\left[-3 \pm 1\right]$
$ = \{-1, -2\}.$
.....
Setting $a = \frac{5}{3}$ yields $\{q,p\}$ of
$ = \frac{1}{2} \times
\left[-\frac{11}{3} \pm \sqrt{\frac{-75}{9} + \frac{40}{3} - 4}\right]$
$ = \frac{1}{2} \times
\left[-\frac{11}{3} \pm \sqrt{\frac{-75 + 120 - 36}{9}}\right]$
$ = \frac{1}{2} \times
\left[-\frac{11}{3} \pm \sqrt{\frac{9}{9}}\right]$
$ = \frac{1}{2} \times
\left[-\frac{11}{3} \pm 1\right]$
$ = \{\frac{-4}{3}, \frac{-7}{3}\}.$
.....
Setting $a = 2$ yields $\{q,p\}$ of
$ = \frac{1}{2} \times
\left[-4 \pm \sqrt{-12 + 16 - 4}\right]$
$ = \frac{1}{2} \times
\left[-4 \pm 0\right]$
$ = \left[-2 \pm 0\right]$
In fact, this means that as $a$ approaches 2, the smaller root will approach (-2) from below and the larger root will approach (-2) from above.
The largest root, $q$ attains the maximum value of $-1$, when $a = 1$.
The smaller root $p$ approaches $-4/3$ from below, as $a$ approaches 2/3.
Addendum
As mathlove indicated in a comment to the original query, the smaller root $p$ never attains a maximum value in the open interval for $a$ of (2/3, 2). I explain this as follows:
First of all, although the boundary points of $a = 2/3$ and $a = 2$ seem to be permissible, the original problem specifies the maximum value for $p$, when it is required that $p < q$. This requirement means that the only permissible values of $a$ are the open interval $(2/3, 2)$ rather than the closed interval $[2/3,2].$
With respect to the value of $a$, if you examine the 4 pertinent points in my answer, (2/3), 1, (5/3), 2, keeping in mind that the only possible values of $a$ that might cause $g'(a) = 0$ are $a=1$ and $a = (5/3)$, you see that :
$E_1:$
as $a$ approaches $2/3$ from above, the smaller root, $p$, approaches -4/3 from below.
$E_2:$
In fact, for no value of $a$ in the open interval $(2/3, 2)$ does $p$ actually $= -4/3.$
Best Answer
If you have a quadratic polynomial $f(x)$ and an exponential function $b(x) = b^x$ where $b>1$, you can show that $b(x)$ surpasses the polynomial by showing that eventually the growth rate of $b(x)$ exceeds the polynomials growth rate.
Since the polynomial's leading term has the biggest effect when $x$ grows very big, that means that the other terms don't matter, so let $f(x) = ax^2$ for an $a>0$. Now, compute the ratio between $f(x+1)$ and $f(x)$: $$ \frac{f(x+1)}{f(x)} = \frac{a(x+1)^2}{ax^2} = \left(\frac{x+1}{x}\right)^2 \, . $$ As you can see, as $x$ grows very big, that ratio between $x+1$ and $x$ grows close to $1$, thus $f(x+1)$ barely increases from $f(x)$ because it is being multiplied by a number close to $1$. Now analyze the ratio between $b(x+1)$ and $b(x)$. By definition, the exponential function multiplies by its base $b$ evey time you increase by $1$, so the ratio between $b(x+1)$ and $b(x)$ is always $b$. However, we stated already that $b>1$. We also found out that $f(x)$ ratio approaches $1$ as $x$ gets really big. Thus, there is a point when $b(x)$ ratio exceeds $f(x)$ ratio, which means that $b(x)$ will start growing faster than $f(x)$ and will eventually outgrow $f(x)$.
Note that I just used terminology like approach and really big because a $7$th grader would not know of limits, so don't nitpick that.
Secondary note: I said $a>0$ and $b>1$ because I assumed that both of the functions would be traveling upwards as you moved right along the $x$-axis.
If you want a downward-facing parabola with $a<0$ and a downward-facing exponential with $0<b<1$, then you can just note that the exponential will just tend towards $0$ when $x$ gets very big, but the quadratic will eventually go below zero if it is facing downwards, thus showing that the exponential will eventually become greater then the quadratic.