What's the other case for finding the missing side of this triangle? Stumped on the second part of this question for a tutoring student. The problem states:

In $\triangle \text{PQR}$, $\text{PQ} = 39 \text{ in.}$, $\text{PR} = 17 \text{ in.}$, and the altitude $\text{PN} = 15 \text{ in.}$ Find $\text{QR}$. Consider all cases.

The first case, which is more obvious, is that $\text{PN}$ and $\text{NQ}$ satisfy the Pythagorean Theorem because $\triangle \text{QPN}$ and $\triangle \text{PNR}$ are right triangles. Therefore:
$15^2+\text{QN}^2=39^2 \implies \text{QN}=36$
$15^2+\text{NR}^2=17^2 \implies \text{NR}=8$
$\text{QR}=\text{QN}+\text{NR}=36+8=44$
The system allows my student to fill in two possible answers. Of course, the first answer is $44$ in., but what is the case that would give the second answer? At first I thought that we could consider the case where $\angle \text{QPR}=90^{\circ}$ and use the Pythagorean Theorem, but I'm pretty sure that would mean that $\text{PN}\neq 15$. What would the other case be?
(keep in mind this student has not covered trig yet)
 A: As @ACB rightfully pointed out, there are two cases. The first case, which I have described in the question, is where $\text{N} \in \overline{\text{QR}}$. The second case is where $\text{N} \notin \overline{\text{QR}}$, $\angle \text{QRP}$ is obtuse, and $\overline{\text{PN}}$ is outside of $\triangle \text{PQR}$ (picture not to scale):

As is pictured, let $x=\text{QR}$ and $y=\text{RN}$.
Using the Pythagorean Theorem, we can create a system of equations to solve $x$ and $y$:
$\Bigg\{\begin{array}[c]
115^2 + (x+y)^2 = 39^2\; (1) \\ 
15^2+y^2=17^2 \; (2)
\end{array}$
Solving equation $(2)$ for $y$, we get $y=8$. Then by substitution we can write equation $(1)$ as:
$\begin{align}
15^2+(x+8)^2&=39^2 \\
225+(x+8)^2&=1521 \\
(x+8)^2&=1296 \\
x+8&=36 \\
x&=28
\end{align}$
Therefore, $x=\text{QR}=28$ and the second answer is $28$ in.
A: Without many words .... ${}{}{}{}{}{}{}{}$

A: Let the base, part $1$, be
$\space QX=\sqrt{QP^2-PX^2}=\sqrt{39^2-15^2}=36.\space$
Let the base, part $2$, be
$\space XR=\sqrt{PR^2-PX^2}
=\sqrt{17^2-15^2}=8.\space$
Then, it follows
$\quad QR=QX+XR=36+8=44$
or
$\quad QR=QX-XR=36-8=28$
