Sum of complex roots of quadratic equation. How to find the sum of the complex roots of the quadratic equation $$px^2+qx+r=0$$ if $r>p>0$, and $p,q,r \in\mathbb R$ ?
 A: The given equation is
$px^2+qx+r=0, \; \text{with} \; r > p > 0 \tag{1}$
and $p, q, r \in \Bbb R$, the real numbers.  If $\rho_1$ and $\rho_2$ are the roots of (1), then they are also roots of
$x^2+\dfrac{q}{p}x+\dfrac{r}{p} = 0, \tag{2}$
which is just (1) divided through by $p \ne 0$.  $\rho_1$ and $\rho_2$ being roots of (2) implies we have the factorization
$x^2+\dfrac{q}{p}x+\dfrac{r}{p} = (x - \rho_1)(x - \rho_2) = x^2 - (\rho_1 + \rho_2)x + \rho_1 \rho_2, \tag{3}$
and if we compare the coefficients of like powers of $x$, i.e., of $x^0 = 1, x^1 = x$ and $x^2$, we see that
$\rho_1 + \rho_2 = -\dfrac{q}{p} \tag{4}$
and
$\rho_1 \rho_2 = \dfrac{r}{p}. \tag{5}$
(4) and (5) hold in the both the cases real and complex roots.  In fact, it is easy to see that they hold over any field, i.e. when $0 \ne p, q, r \in \Bbb F$ an arbitrary field in which (1) has a root.
Note Added in Edit: Returning to the case $\Bbb F = \Bbb R$, $r >p > 0$, with $\rho_1, \rho_2 \in \Bbb C$ but $\rho_1, \rho_2 \notin \Bbb R$, note that in this event we have $\rho_2 = \bar {\rho}_1$, i.e. $\rho_1$ and $\rho_2$ are complex conjugates.Thus by (5),
$\vert \rho_1 \vert^2 = \rho_1 \bar{\rho_1} = \rho_1 \rho_2 = \dfrac{r}{p}, \tag{6}$
and likewise
$\vert \rho_2 \vert^2 = \dfrac{r}{p}, \tag{7}$
so that
$\vert \rho_1 \vert = \vert \rho_2 \vert = \sqrt{\dfrac{r}{p}}. \tag{8}$
Thus
$\vert \rho_1 \vert + \vert \rho_2 \vert = 2\sqrt{\dfrac{r}{p}} > 2 \tag{9}$
since $r > p$.  This is option (d) in user140013's comment.  End of Note.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: $$
x = \frac{-q \pm \sqrt{q^2 - 4pr}}{2p}
$$
Regardless:
$$
\frac{-q + \sqrt{q^2 - 4pr}}{2p} + \frac{-q - \sqrt{q^2 - 4pr}}{2p} =\frac{-2q + \sqrt{q^2 - 4pr} - \sqrt{q^2 - 4pr}}{2p}= \frac{-2q + 0}{2p} = -\frac{q}{p}
$$
edit
OK, I think I understand the problem now.  For there to be complex roots, we have:
$$
q^2 < 4pr \rightarrow |q| < 2\sqrt{pr}
$$
We need to handle two different cases:


*

* q > 0
Here we have $-\frac{q}{p} = -\frac{|q|}{p}$.  Since $|q| < 2\sqrt{pr}$, this gives a bound from a "small" negative number.  The actual $q$ must be smaller, and thus must be a smaller negative which is really a larger number (a more positive number), thus we get:
$$
\text{sum} > -2\frac{\sqrt{pr}}{p}\rightarrow \text{sum} > -2\sqrt{\frac{r}{p}}
$$
Since $r > p$, $\sqrt{\frac{r}{p}} > 1$.  This gives a minimum (negative) number of $-2$.  Since the sum should be greater, this means we indeed have $\text{sum} > -2$.

*$q \leq 0$
In this case $-\frac{q}{p} = \frac{|q|}{p}$.  Now we need a legitimately smaller number (smaller positive number):
$$
\text{sum} < 2\sqrt{\frac{r}{p}} < 2
$$


This would suggest that the sum is between $-2$ and $+2$.
