The maximum value of the smaller root of given quadratic function

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 $a = frac{1}{6} times left[ 8 pm 2right].$

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 1right]$

$ = {-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 1right]$

$ = {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 0right]$

$ = left[-2 pm 0right]$

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 $= -4/3.$