Question 68: Given $a, b \in R$, such that the equation $x^{4}+a x^{3}+2 x^{2}+b x+1=0$ has real roots, find the minimum possible value of $a^{2}+b^{2}$.

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Question 68, Solution: It is evident that the real root of the equation is not 0. According to the given conditions:
$$
\mathrm{ax}+\frac{\mathrm{b}}{\mathrm{x}}=-\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{2}
$$

Let $t=x+\frac{1}{x}$, then $|t| \geq 2$. According to the Cauchy-Schwarz inequality, and combining with (1), we have:
$$
\begin{array}{l}
\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)\left(\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}\right) \geq\left(\mathrm{ax}+\frac{\mathrm{b}}{\mathrm{x}}\right)^{2}=\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{4} \\
\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2} \geq \frac{\mathrm{t}^{4}}{\mathrm{t}^{2}-2}=\left(\mathrm{t}^{2}-2\right)+\frac{4}{\mathrm{t}^{2}-2}+4 \geq 4+4=8 \\
\end{array}
$$

On the other hand, when $a=b=-2$, the equation $x^{4}+a x^{3}+2 x^{2}+b x+1=0$ has a real root $x=1$, and $a^{2}+b^{2}=8$.