Calculating the Discriminant for The Following Quadratic: Consider $$ p_{k+1} \zeta_{k+1}^2 + q_{k+1}\zeta_{k+1} + r_{k+1} = 0 $$
where 
$$p_{k+1} = p_0P_k^2 + q_0P_kQ_k + r_0Q_k^2$$
$$q_{k+1} = 2p_0P_kP_{k-1} + q_0 ( P_kQ_{k-1} + Q_kP_{k-1} ) + 2r_0Q_kQ_{k-1}$$
$$r_{k+1} = p_0P_{k-1}^2  + q_0 P_{k-1}Q_{k-1} + r_0Q_{k-1}^2 = p_k $$
Given that $p_k \neq 0$ for any given $k$ 
Now the discriminant of this equation turns out to be: 
$$\Delta = (q_0^2 - 4p_0r_0 )( P_kQ_{k-1} - P_{k-1}Q_k)^2 $$ 
Specifically, $$C_k = \frac{ P_k } { Q_k } $$ denote the $k-th$ convergent of a simple continued fraction, so the identity $$P_kQ_{k-1}+Q_kP_{k-1} = \pm 1 $$ may be useful for our calculation. 
I noticed that the first bracketed term of $\Delta$ was simply when $k=-1$, so is there some sort of easy way of achieving that expression without any intense algebra? 
Kind Regards, 
 A: Let's see, for you, $\alpha = P_k,\gamma = Q_k, $ $\beta = P_{k-1},\delta = Q_{k-1}, $
$$
p =
\left(
\begin{array}{cc}
P_k &  P_{k-1} \\
Q_k &  Q_{k-1}
\end{array}
\right)
$$
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We have a binary form $\langle A,B,C \rangle$ meaning $f(x,y) = A x^2 + B xy + C y^2.$ We create the Hessian matrix
$$
h =
\left(
\begin{array}{cc}
2A & B \\
B & 2 C
\end{array}
\right)
$$
with discriminant
$$ \Delta = B^2 - 4 AC. $$
To get back to the triple of coefficients we halve the diagonal entries but keep one of the off diagonal entries as is, for $B.$
Given a matrix
$$
p =
\left(
\begin{array}{cc}
\alpha & \beta \\
\gamma & \delta
\end{array}
\right)
$$
we calculate the symmetric matrix $p^T h p$ to give a new one,
$$ \langle A \alpha^2 + B \alpha \gamma + C \gamma^2, 2A \alpha \beta + B(\alpha \delta + \beta \gamma) + 2 C \gamma \delta, A \beta^2 + B \beta \delta + C \delta^2  \rangle $$ 
Note that, about the discriminant $\Delta,$ we automatically have the new discriminant being $\Delta \det^2 p.$
Given
$$
Q =
\left(
\begin{array}{ccc}
\alpha^2 &  2 \alpha \beta & \beta^2 \\
\alpha \gamma & \alpha \delta + \beta \gamma & \beta \delta \\
\gamma^2 & 2 \gamma \delta & \delta^2
\end{array}
\right),
$$
there is a typographical error on  page 23 of Magnus, actually $\det Q = (\alpha \delta - \beta \gamma)^3.$
If we now write the triple $(A,B,C)$ as a row vector, we find
$$
(A,B,C)
\left(
\begin{array}{ccc}
\alpha^2 &  2 \alpha \beta & \beta^2 \\
\alpha \gamma & \alpha \delta + \beta \gamma & \beta \delta \\
\gamma^2 & 2 \gamma \delta & \delta^2
\end{array}
\right) = 
$$
$$ 
( A \alpha^2 + B \alpha \gamma + C \gamma^2, 2A \alpha \beta + B(\alpha \delta + \beta \gamma) + 2 C \gamma \delta, A \beta^2 + B \beta \delta + C \delta^2) 
$$
Compare our earlier $ p^T h p =$
$$ \langle A \alpha^2 + B \alpha \gamma + C \gamma^2, 2A \alpha \beta + B(\alpha \delta + \beta \gamma) + 2 C \gamma \delta, A \beta^2 + B \beta \delta + C \delta^2  \rangle $$
