Therefore we have $\lambda_1<\lambda_2$.

Therefore we have the following

\[thm:cuts\] Let $\Delta$ be the discriminant of the quartic polynomial $$\label{eq:quartic}
\begin{split}
a^2(1-c)\xi^4
&+2(a^2(1-c\beta)+a(1-c(1-\beta))\xi^3\\
&+(1-c(1-\beta)+a^2(1-c\beta)+4a)\xi^2 +2(1+a)\xi+1=0.
\end{split}$$ then if $\Delta<0$, the support of $\underline{F}$ consists of a single interval.

We can treat (\[eq:curve2\]) as a polynomial in $\xi$ then $\lambda_k$, $k=1,\ldots,4$ are the zeroes of its discriminant $D_3(z)=(az)^4\prod_{i<j}(\xi_i-\xi_j)^2$. $$\label{eq:D3}
\begin{split}
D_3(z)&=(1-a)^2z^4+(2A_2^2B_1+2A_2B_2-4A_2^3-12aB_1+18aA_2)z^3\\
&+(B_2^2+A_2^2B_1^2+4A_2B_1B_2-12A_2^2B_2-12aB_1^2+18aB_2+18aA_2B_1-27a^2)z^2\\
&+(2B_1B_2^2+2A_2B_2B_1^2-12A_2B_2^2-4B_1^3a+18aB_1B_2)z+B_1^2B_2^2-4B_2^3.
\end{split}$$ The zeros of (\[eq:D3\]) then correspond to the branch points of the Riemann surface ${\mathcal{L}}$. These branch points are given on ${\mathcal{L}}$ by $(\lambda_k,\gamma_k)$, for $k=1,\ldots,4$.

Since the leading coefficient of $D_3(z)$ is $(1-a)^2>0$, we see that the sign of $D_3(z)$ and hence the 3 roots of the cubic polynomial (\[eq:curve2\]) behave as follows for $z\in\mathbb{R}$. $$\label{eq:realim}
\begin{split}
&1. \quad z\in\mathbb{R}\setminus [\lambda_1,\lambda_2],\quad
D_3(z)>0\Rightarrow
\textrm{$\xi$ has 3 distinct real roots} \\
&2. \quad z\in (\lambda_1,\lambda_2),\quad
D_3(z)<0\Rightarrow\textrm{$\xi$ has 1 real and 2 complex roots}.
\end{split}$$ In particular, since the roots coincide at the branch points, the $\gamma_k$, $k=1,\ldots,4$ is the values of the double root of the cubic (\[eq:curve2\]) when $z=\lambda_k$.

We can now compute the probability density $\underline{F}$.

\[thm:density\] Suppose $\Delta<0$. Then the p.d.f $\underline{F}$ is supported on $[\lambda_1,\lambda_2]$ with the following density $d\underline{F}(z)=\rho(z)dz$ $$\label{eq:density}
\frac{3}{2\pi}\left|\left(\frac{r(z)+\sqrt{-\frac{1}{27a^4z^4}D_3(z)}}{2}\right)^{\frac{1}{3}}-\left(\frac{r(z)-\sqrt{-\frac{1}{27a^4z^4}D_3(z)}}{2}\right)^{\frac{1}{3}}\right|,$$ where $D_3(z)$ 