Answer
True.
Work Step by Step
If there exists an x-intercept,
the y-coordinate of that point must be zero.
If $y=0,$
$x^{2}-2=0\qquad/$+$2$
$x^{2}=+2$
$x=+\sqrt{2}$ or $x=-\sqrt{2}.$
So, there are two x-intercepts,
$(-\sqrt{2},0)$ and $(\sqrt{2},0)$
(and no other)
