215. Given the quadratic equation $x^{2}+p x+q=0$. Find the equation whose roots: 1) differ from the roots of the given equation only in sign, 2) are the reciprocals of the roots of the given equation.
Solution. Let the roots of the given equation be $x_{1}, x_{2}$, then, as is known,

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

If the roots of the desired equation are ( $-x_{1}$ ) and ( $-x_{2}$ ), then

$$
\begin{gathered}
\left(-x_{1}\right)+\left(-x_{2}\right)=-\left(x_{1}+x_{2}\right)=-(-p)=p \\
\left(-x_{1}\right)\left(-x_{2}\right)=x_{1} x_{2}=q
\end{gathered}
$$

Thus, the desired equation will have the form: $x^{2}-p x+q=0$.

If the roots of the desired equation are the reciprocals of the roots of the given quadratic equation, then they will be: $\frac{1}{x_{1}}$ and $\frac{1}{x_{2}}$. The sum and product of these roots will be expressed as:

$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}=-\frac{p}{q} ; \frac{1}{x_{1}} \cdot \frac{1}{x_{2}}=\frac{1}{q}
$$

The desired equation will have the form:

$$
x^{2}+\frac{p}{q} x+\frac{1}{q}=0 \text { or } q x^{2}+p x+1=0 .
$$