If $x^3 + 5x^2 + px + q = 0$ and $x^3 + 7x^2 + px + r = 0$ have two common roots then there must exists a common monic binomial factor, $F(x)$.
So $F(x)$ must be a divisor of
$$(x^3 + 7x^2 + px + r) - (x^3 + 5x^2 + px + q) = 2x^2 - (q - r) $$
So $F(x) = x^2 - \dfrac 12(q - r)$.
So, for some $s$
\begin{align}
x^3 + 7x^2 + px + r
&= F(x)(x-s) \\
&= \left[x^2 - \dfrac 12(q - r)\right](x-s)\\
&= x^3 - sx^2 - \dfrac 12(q - r)x + \dfrac 12(q - r)s\\
\hline
s &= -7 \\
p &= -\dfrac 12(q - r) \\
r &= -\dfrac 72(q - r) \\
\hline
s &= -7 \\
p &= \dfrac 15 q \\
r &= \dfrac 75 q
\end{align}
And, for some $t$
\begin{align}
x^3 + 5x^2 + px + q
&= \left[x^2 - \dfrac 12(q - r)\right](x-t)\\
&= x^3 - tx^2 - \dfrac 12(q - r)x + \dfrac 12(q - r)t\\
\hline
t &= -5 \\
p &= -\dfrac 12(q - r) \\
q &= -\dfrac 52(q - r) \\
\hline
t &= -5 \\
p &= \dfrac 15 q \\
r &= \dfrac 75 q
\end{align}
So the third roots are $-7$ and $-5$.
Reality check:
So, suppose $p = -\dfrac 15 q, r = \dfrac 75 q$. To get rid of the fractions, lets let q = 10k.
Then
- $p = 2k$
- $r = 14k$
- $q = 10k$
- $s = -7$
- $t = -5$
- $F(x) = x^2 + 2k$
- $x^3 + 7x^2 + px + r = x^3 + 7x^2 + 2kx + 14k$
- $x^3 + 5x^2 + px + q = x^3 + 5x^2 + 2kx + 10k$
- $F(x)(x-s) = (x^2 + 2k)(x+7) = x^3 + 7x^2 + 2kx + 14k$
- $F(x)(x-t) = (x^2 + 2k)(x+5) = x^3 + 5x^2 + 2kx + 10k$