Let $f(x) \equiv x^{2}+p x+q=0$ be an equation with roots $x_{1}$ and $x_{2}$, which are numbers different from 1. Which quadratic equation has the roots

$$
y_{1}=\frac{x_{1}+1}{x_{1}-1} \quad \text { and } \quad y_{2}=\frac{x_{2}+1}{x_{2}-1} \text {. }
$$
I. Solution. Let

$$
y=\frac{x+1}{x-1}
$$

From this,

$$
x=\frac{y+1}{y-1}
$$

Substitute this expression for $x$ into the equation $f(x)=0$; after removing the fractions, we get:

$$
f(y) \equiv(y+1)^{2}+p(y+1)(y-1)+q(y-1)^{2}=0
$$

or

$$
f(y) \equiv(1+p+q) y^{2}+2(1-q) y+1-p+q=0
$$

II. Solution. Determine the values of $y_{1}+y_{2}$ and $y_{1} y_{2}$ using the coefficients of the equation $f(x)=0$.

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

Therefore, $y_{1}$ and $y_{2}$ are the roots of the equation

$$
y^{2}+\frac{2(1-q)}{q+p+1} y+\frac{q-p+1}{q+p+1}=0
$$

or

$$
f(y) \equiv(1+p+q) y^{2}+2(1-q) y+1-p+q=0
$$

In the case of $p+q+1=0$, we have $p=-(q+1)$, and thus the roots of

$$
f(x) \equiv x^{2}-(q+1) x+q=0
$$

are $q$ and $1$.

However, this case is excluded when one of the roots of $f(x)=0$ is 1.

Sinay Marianne (Ref. girl's high school VI. grade, Miskolc)