4.97 Real number $p \geqslant \frac{1}{4}$. Find all positive real solutions $x$ of the following equation:
$$
\log _{\sqrt{2}}^{2} x+2 \log _{\sqrt{2}} x+2 \log _{\sqrt{2}}\left(x^{2}+p\right)+p+\frac{15}{4}=0 \text {. }
$$
【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,
the left side of the original equation $>\log _{\sqrt{2}}^{2} x+2 \log _{\sqrt{2}} x+2 \log _{\sqrt{2}} x+\frac{1}{4}+\frac{15}{4}$
$$
\begin{array}{l}
=\log _{\sqrt{2}}^{2} x+4 \log _{\sqrt{2}} x+4 \\
=\left(\log _{\sqrt{2}} x+2\right)^{2} \\
\geqslant 0,
\end{array}
$$

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

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

That is,
$$
x=\frac{1}{2}.
$$

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 $x=\frac{1}{2}$.