225. Find the condition under which the difference of the roots in the equation $x^{2}+p x+$ $+q=0$ would be equal to $a$.
Solution. Let the roots of the given equation be denoted by $x_{1}$ and $x_{2}$, then

$$
x_{1}+x_{2}=-p, \quad x_{1} x_{2}=q, \quad x_{1}-x_{2}=a
$$

From the first and third equations, we find $x_{1}$ and $x_{2}$, and substituting the found values for $x_{1}$ and $x_{2}$ into the second equation, we get:

$$
x_{1}=\frac{a-p}{2}, x_{2}=-\frac{a+p}{2}
$$

and

$$
x_{1} \cdot x_{2}=\frac{p^{2}-a^{2}}{4}=q
$$

i.e.,

$$
a^{2}-p^{2}=-4 q
$$