The roots of the equations $f(x)=0$ and $f(y)=0$ coincide if the ratios of the corresponding coefficients are equal, that is,

$$
\frac{1+p+q}{1}=\frac{2(1-q)}{p}=\frac{1-p+q}{q} \cdots
$$

If these ratios are equal, then their value is equal to the ratio of the sum of the two extreme numerators and the sum of the denominators, and this is

$$
\frac{1+p+q+1-p+q}{1+q}=\frac{2(1+q)}{1+q}=2 \quad \text { provided } \quad q \neq 1
$$

According to these: $\quad 1+p+q=2$, and from (1) $1-q=p, \quad 1-p=q$.

All three say the same thing, i.e., $\quad p+q=1 \ldots$

Then indeed: $1+(p+q)=2 ; \quad 2(1-q)=2 p$

and

$$
1-p+q=p+q-p+q=2 q
$$

that is, the coefficients of the $f(y)=0$ equation are twice those of the corresponding coefficients of the $f(x)=0$ equation.

Let's also examine the case where $q=-1$. Then from the ratios below (1):

$$
\frac{p}{1}=\frac{4}{p}=\frac{-p}{-1}
$$

From this

$$
p^{2}=4, \quad \text { that is } \quad p=+2
$$

For $p=+2 \quad$ we have $\quad f_{1}(x) \equiv x^{2}+2 x-1=0 \quad$ and $\quad f_{1}(y) \equiv 2 y^{2}+4 y-2=0$

For $p=-2 \quad$ we have $\quad f_{2}(x) \equiv x^{2}-2 x-1=0 \quad$ and $\quad f_{2}(y) \equiv-2 y^{2}+4 y+2=0$.

It can be seen that in the case of $q=-1, p= \pm 2$ as well, the

$$
f_{1}(x)=0 \quad \text { and } \quad f_{1}(y)=0 \quad \text { or } \quad f_{2}(x)=0 \quad \text { and } \quad f_{2}(y)=0
$$

equations have the same roots. The case $p=+2, q=-1$ fits into the general $p+q=1$ relationship.
Let the result of the previous task be denoted by $f(y)=0$. Determine the values of $p$ and $q$ for which the roots of the equation $f(y)=0$ coincide with the roots of the equation $f(x)=0$!