5-68 Find all quadratic trinomials $p(x)$, which take a minimum value of $-\frac{49}{4}$ at $x=\frac{1}{2}$, and the sum of the fourth powers of its two roots is 337.
[Solution] From the given, we can set
$$
p(x)=a\left(x-\frac{1}{2}\right)^{2}-\frac{49}{4} \quad(a>0),
$$

i.e., $p(x)=a x^{2}-a x+\frac{a-49}{4}$.
Let the two roots of $p(x)$ be $x_{1}, x_{2}$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=1, \\
x_{1} x_{2}=\frac{a-49}{4 a} .
\end{array}\right.
$$

But $\quad x_{1}^{4}+x_{2}^{4}$
$=\left(x_{1}+x_{2}\right)^{4}-4 x_{1} x_{1}\left(x_{1}^{2}+x_{2}^{2}\right)-6 x_{1}^{2} x_{2}^{2}$
$=\left(x_{1}+x_{2}\right)^{4}-4 x_{1} x_{2}\left[\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right]-6 x_{1}^{2} x_{2}^{2}$,
$\therefore \quad 337=1^{4}-\frac{a-49}{a} \cdot\left(1-\frac{a-49}{2 a}\right)-6 \cdot \frac{(a-49)^{2}}{16 a^{2}}$,
$336=-\frac{a-49}{a}+\frac{(a-49)^{2}}{2 a^{2}}-\frac{3(a-49)^{2}}{8 a^{2}}$,
$$
336=-\frac{a-49}{a}+\frac{(a-49)^{2}}{8 a^{2}} .
$$

Then $8 \times 336 a^{2}=-8 a(a-49)+(a-49)^{2}$,
Rearranging gives $2695 a^{2}-294 a-2401=0$.
So $a_{1}=1, a_{2}=-\frac{2401}{2695}$ (discard)
Therefore, $\quad p(x)=x^{2}-x-12$.