8. Find the minimum value of the discriminant of a quadratic trinomial, the graph of which has no common points with the regions located below the x-axis and above the graph of the function $y=\frac{1}{\sqrt{1-x^{2}}}$.
Answer: -4.

Solution. Since the discriminant of a quadratic trinomial

$$
D=b^{2}-4 a c
$$

we will find the maximum value of the product $a c$. Since for all $x$ the inequality $a x^{2}+b x+c \geqslant 0$ holds, then $a \geqslant 0, c \geqslant 0$. Let

$$
f(x)=\frac{1}{\sqrt{1-x^{2}}}-\left(a x^{2}+b x+c\right)
$$

Since for all $|x|<1$ it is true that $f(x) \geqslant 0$, then

$$
0 \leqslant f\left(\frac{1}{\sqrt{2}}\right)+f\left(-\frac{1}{\sqrt{2}}\right)=2 \sqrt{2}-(a+2 c)
$$

Thus, $a+2 c \leqslant 2 \sqrt{2}$, so from the inequality between the arithmetic mean and the geometric mean, we get:

$$
2 \sqrt{2} \geqslant a+2 c \geqslant 2 \cdot \sqrt{a \cdot 2 c}=2 \sqrt{2} \sqrt{a c}
$$

Therefore, $a c \leqslant 1$.

We will prove that 1 is the maximum of $a c$. Indeed, let $a=2 c=\sqrt{2}, b=0$, then the original inequality takes the form:

$$
\sqrt{2} x^{2}+\frac{1}{\sqrt{2}} \leqslant \frac{1}{\sqrt{1-x^{2}}} \Leftrightarrow\left\{\begin{array}{l}
\left(2 x^{2}-1\right)^{2}\left(x^{2}+1\right) \geqslant 0 \\
|x|<1
\end{array}\right.
$$

Therefore, for the discriminant of the quadratic trinomial, the following inequalities hold:

$$
D=b^{2}-4 a c \geqslant-4 a c \geqslant-4
$$

and its minimum value is -4.