10. If the curve $y=\left|x^{2}-3\right|$ intersects the line $y=2 x+$ $k$ at exactly three points, then the value of $k$ is $\qquad$
$10.4,2 \sqrt{3}$.
(1) If the line $y=2 x+k$ is tangent to the curve $y=-\left(x^{2}-3\right)$ at point $P$ (figure omitted), then $2 x+k=3-x^{2}$, which means $x^{2}+2 x+k-3=0$ has a discriminant of 0. Thus,
$$
2^{2}=4(k-3) \text {. }
$$

Solving for $k$ gives $k=4$.
(2) If the line $y=2 x+k$ passes through $Q(-\sqrt{3}, 0)$, then $0=-2 \sqrt{3}+k$. In this case, $k=2 \sqrt{3}$.