This is a question I found in a book.
Let $0<p<q$ and $a\neq0$ such that the equation $$px^2+4\lambda xy+qy^2+4a\left(x+y+1\right)=0$$ represents a pair of straight lines, then $a$ can lie in the interval
(A) $\left(-\infty ,\infty\right)$
(B) $\left(-\infty ,p\right]$
(C) $\left[p,q\right]$
(D) $\left[q,\infty\right]$
And this is the solution provided. The answer is given to be options (B) and (D).
$p~.~q~.~4a+2~.~2a~.~2a~.~2\lambda-p~.~4a^2-q~.~4a^2-4a~.~4\lambda^2=0$
$\Rightarrow~ 4\lambda^2-4a\lambda+\{\left(p+q\right)a-pq\}=0~~~\left(\because a\neq0\right)$
$\because~\lambda\in\mathbb{R},~16a^2-4.4\{\left(p+q\right)a-pq\}\geq0$
or $[a-p\left(a-q\right)]\geq0$
$\therefore a\leq p~\text{or}~a\geq q$
I understood how they went with the solution; for the general equation of a conic $ax^2+2hxy+by^2+2gx+2fy+c=0$ to be an equation of a pair of straight lines, the necessary and sufficient condition is $$\Delta=\left\lvert\begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix}\right\rvert=0$$
What I don't understand is how they went from the third line to the fourth line; and then from the fourth line to the fifth line.
After taking out the factor of $16$, the condition is $a^2-(p+q)a+pq=(a-p)(a-q)\geqslant0$, which means that either $a\leqslant p$ or $a\geqslant q$. The fourth line appears to be misprinted.