4.97 Real number $p \geqslant \frac{1}{4}$. Find all positive real solutions $x$ of the following equation:
$$\log _{\sqrt{2}}^{2} x+2 \log _{\sqrt{2}} x+2 \log _{\sqrt{2}}\left(x^{2}+p\right)+p+\frac{15}{4}=0$$
［Solution］ When $p>\frac{1}{4}$,
$$\left(x-\frac{1}{2}\right)^{2}+p-\frac{1}{4}>0$$

That is,
$$x^{2}+p>x$$

Since $\sqrt{2}>1, x>0$, we have
$$\log _{\sqrt{2}}\left(x^{2}+p\right)>\log _{\sqrt{2}} x$$

Thus,
$$\begin{aligned}
\text { the left side of the original equation } & >\log _{\sqrt{2}}^{2} x+2 \log _{\sqrt{2}} x+2 \log _{\sqrt{2}} x+\frac{1}{4}+\frac{15}{4} \\
& =\log _{\sqrt{2}}^{2} x+4 \log _{\sqrt{2}} x+4 \\
& =\left(\log _{\sqrt{2}} x+2\right)^{2} \\
& \geqslant 0
\end{aligned}$$

Therefore, when $p>\frac{1}{4}$, the original equation has no solution.
When $p=\frac{1}{4}$, from $\left(x-\frac{1}{2}\right)^{2} \geqslant 0$, we get
$$x^{2}+\frac{1}{4} \geqslant x$$

Thus,
$$\log _{\sqrt{2}}\left(x^{2}+\frac{1}{4}\right) \geqslant \log _{\sqrt{2}} x$$

At this point,
$$\begin{aligned}
\text { the left side of the equation } & =\log _{\sqrt{2}}^{2} x+2 \log _{\sqrt{2}} x+2 \log _{\sqrt{2}}\left(x^{2}+\frac{1}{4}\right)+4 \\
& \geqslant \log _{\sqrt{2}}^{2} x+2 \log _{\sqrt{2}} x+2 \log _{\sqrt{2}} x+4 \\
& =\left(\log _{\sqrt{2}} x+2\right)^{2} \\
& \geqslant 0
\end{aligned}$$

Therefore, the necessary and sufficient condition for the original equation to have a positive real solution is
$$x-\frac{1}{2}=0 \text { and } \log _{\sqrt{2}} x+2=0 \text {, }$$

That is $\square$
$$x=\frac{1}{2}$$

In summary, when $p>\frac{1}{4}$, the original equation has no positive real solution; when $p=\frac{1}{4}$, the original equation has only one positive real solution $x=\frac{1}{2}$.