Answer
See below
Work Step by Step
The pyramid's volume is given by: $V=\frac{1}{3}Bh$
where $B=x^2$ is the area of its base.
$h=2x-5$ is the height
Obtain: $V=\frac{1}{3}x^2(2x-5)=3\\
\rightarrow x^2(2x-5)=9\\
\rightarrow 2x^3-5x^2=9\\
\rightarrow 2x^3-5x^2-9=0$
The leading coefficients: $\pm 1,\pm 2$
The constant terms: $\pm 1, \pm 3, \pm 9$
The possible rational zeros are: $\pm\frac{1}{1},\pm \frac{3}{1},\pm \frac{9}{1},\pm\frac{1}{2},\pm \frac{3}{2},\pm \frac{9}{2}$
The values must be positive. Hence, the possible solutions of the problem are: $3,9,\frac{9}{2}$
