3. Let $a>0$, then the equation $\sqrt{a-x^{2}}=\sqrt{2}-|x|$ has unequal real roots. Therefore, the range of values for $a$ is ().
(A) $a>0$
(B) $0<a<1$
(C) $a=1$
(D) $a \geqslant 1$
3. D.

Square both sides of the equation, we get
$$
a-x^{2}=2+x^{2}-2 \sqrt{2}|x| \text {. }
$$

Rearrange and square again, we get $\left(a-2-2 x^{2}\right)^{2}=8 x^{2}$,
simplify to $\quad 4 x^{4}-4 a x^{2}+a^{2}-4 a+4=0$.
It is easy to see that, when $x \neq 0$, if $x$ is a root of the original equation, then $-x$ is also a root of the original equation, so we only need to ensure that the discriminant of (*) $\geqslant 0$ :
$$
\Delta=16 a^{2}-16\left(a^{2}-4 a+4\right) \geqslant 0 \text {, which gives } a \geqslant 1 \text {. }
$$

If $x=0$ is a root, it must be that $a=2$, in which case the original equation also has real roots $\pm \sqrt{2}$, which still satisfies the condition $a \geqslant 1$. (This solution is provided by the author)