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
$$p(x)=a\left(x-\frac{1}{2}\right)^{2}-\frac{49}{4} \quad(a>0)$$

which means $p(x)=a x^{2}-a x+\frac{a-49}{4}$.
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 $\square$
$$\begin{aligned}
& x_{1}^{4}+x_{2}^{4} \\
= & \left(x_{1}+x_{2}\right)^{4}-4 x_{1} x_{1}\left(x_{1}^{2}+x_{2}^{2}\right)-6 x_{1}^{2} x_{2}^{2} \\
= & \left(x_{1}+x_{2}\right)^{4}-4 x_{1} x_{2}\left[\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right]-6 x_{1}^{2} x_{2}^{2},
\end{aligned}$$
$\therefore \quad 337=1^{4}-\frac{a-49}{a} \cdot\left(1-\frac{a-49}{2 a}\right)-6 \cdot \frac{(a-49)^{2}}{16 a^{2}}$,
$$\begin{array}{l}
336=-\frac{a-49}{a}+\frac{(a-49)^{2}}{2 a^{2}}-\frac{3(a-49)^{2}}{8 a^{2}}, \\
336=-\frac{a-49}{a}+\frac{(a-49)^{2}}{8 a^{2}} .
\end{array}$$

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