2.79 Find the coefficients of the equation $x^{2}+p x+q=0$ given that the difference between the roots of the equation is 5, and the difference between their cubes is 35.
2.79 We have $x_{1}-x_{2}=5, x_{1}^{3}-x_{2}^{3}=35$, where $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}+p x+q=0$. Moreover,

$$
x_{1}=\frac{-p+\sqrt{p^{2}-4 q}}{2} \text { and } x_{2}=\frac{-p-\sqrt{p^{2}-4 q}}{2}
$$

Thus $x_{1}^{3}-x_{2}^{3}=\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)$, or

$$
5\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)=35, \text { or } x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=7
$$

i.e., we obtain the system

$\left\{\begin{array}{l}p^{2}-q=7 \\ \sqrt{p^{2}-4 q}=5 .\end{array}\right.$

Solving it, we find $p=1, q=-6$ and $p=-1, q=-6$.

Answer: $p=1, q=-6 ; p=-1, q=-6$.