Problem source

Consider the quadratic equation
$$ 
nx^2+2x  \sqrt{pn^2+q} + rn + s = 0, 
\tag{*}
$$ 
where $p>0$, $p\neq r$ and $n=1$, $2$, $3$, $\ldots$ .
\begin{questionparts}
\item For the case where $p=3$, $q=50$, $r=2$, $s=15$, 
find the set 
of values of $n$ for which  equation $(*)$ has no real roots.
\item Prove that if $p < r$ and $4q(p-r) > s^2$, then $(*)$
has no real roots for any value of $n$.
\item If $n=1$, $p-r=1$  and $q={s^2}/8$, 
show that $(*)$ has real roots if, and only if, 
$s \le 4-2\sqrt{2}\ $ or $\ s \ge 4+2\sqrt{2}$.
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$ \begin{align*}
&& 0 &= nx^2 + 2\sqrt{3n^2+50}x + 2n + 15 \\
&& 0 &> \Delta = 4(3n^2+5) - 4\cdot n \cdot (2n + 15) \\
\Leftrightarrow && 0 &> n^2-15n+5\\
\text{cv}: && n &= \frac{15 \pm \sqrt{225 - 20}}{2} \\
&&&\approx \frac{15\pm14.x}{2}\\
\Leftrightarrow &&n &\in \{1, 2, \cdots, 14\}
\end{align*}

\item $\,$ \begin{align*}
&& 0 &> \Delta = 4(pn^2+q) - 4\cdot n \cdot (rn+s) \\
\Leftrightarrow && 0&>(p-r)n^2-sn+q
\end{align*}

Which is always true if $r > p$ and $s^2 < 4q(p-r)$

\item $\,$ \begin{align*}
&& 0 &= x^2 + 2\sqrt{p+q}x+ r+s \\
&& 0 &\leq \Delta = 4(p+q) - 4(r+s) \\
&& 0 &\leq 1 + s^2/8 - s \\
\text{c.v}: && s &= \frac{1 \pm \sqrt{1-4 \cdot \frac{1}{8}}}{2\cdot\frac18} \\
&&&= 4 \pm 4\sqrt{\frac12} \\
&&&= 4 \pm 2\sqrt{2} \\
\Rightarrow && s \leq 4 - 2\sqrt{2} &\text{ or } s \geq 4 + 2\sqrt{2}
\end{align*}
\end{questionparts}