Compare the quadratic equation for the purely periodic continued fraction with periodic terms $q_0, q_1, . . . , q_n$ to the quadratic equation for the purely periodic continued fraction with periodic terms $q_n, . . . , q_1, q_0$ (i.e. in reversed order)
For the periodic terms $q_0, q_1, . . . , q_n$, the continued fraction has the form: $$x={q_0+\cfrac{1}{q_1+\cfrac{1}{q_n+\cfrac{1}{{x}}}}}$$ and has as quadratic equation $x^2.B_n + x.(B_{n-1}-A_n) - A_{n-1} = 0$, where $x = \frac{A_{n-1}+x.A_n}{B_{n-1}+x.B_n}$
For the periodic terms $q_n, ...,q_1,q_0$, the continued fraction has the form:
$$x={q_n+\cfrac{1}{q_{n-1}+\cfrac{1}{q_0+\cfrac{1}{{x}}}}}$$ and has as quadratic equation (as I suppose, I'm not sure) $x^2-x.q_0-1=0$, where $x=\frac{A_1}{B_1} = \frac{1+x.q_0}{x}$
So if I compare these two quadratic equation, I find that they are equal for $n=0$ and $x^2.B_n + x.(B_{n-1}-A_n) - A_{n-1}$ $>$ $x^2-x.q_0-1=0$ for $n>0$
Is my answer correct? Please I really need help in solving this