Answer
$x\in \{ -3, -\frac{1}{2}, \frac{1}{2}, 3\}$
Work Step by Step
$g(x)=4x^4-37x^2+9,$ Lets let $x^2=k$
$=4k^2-37k+9,$
$=4k^2-k-36k+9,$
$=k(4k-1)-9(4k-1),$
$=(k-9)(4k-1),$
$=(x^2-9)(4x^2-1),$
$=(x-3)(x+3)(2x-1)(2x+1),$
$x\in \{ -3, -\frac{1}{2}, \frac{1}{2}, 3\}$
x-intercept of the graph is where the graph crosses $x-axis$. Thus,
$x\in \{ -3, -\frac{1}{2}, \frac{1}{2}, 3\}$ is the x-intercept of the graph.
