(10. The equation $x^{2}+a|x|+a^{2}-3=0(a \in \mathbf{R})$ about $x$ has a unique real solution, then $a=$ $\qquad$ .
$10 f(x)=x^{2}+a|x|+a^{2}-3$ is an even function, the only real solution must be 0, so $a^{2}-3=0$, getting $a= \pm \sqrt{3}$, it is easy to verify that $a=-\sqrt{3}$ does not meet the condition, hence $a=\sqrt{3}$.