jmc

The distance between the origin and $F_1$ is 2, and the distance between the origin and $F_2$ is 3, so every point $P$ on the ellipse satisfies $$P{F}_1+P{F}_2=5.$$So, if $(x,0)$ is an intercept of the ellipse, then $$\sqrt{x^2+4}+\sqrt{(x-3{)}^2}=5.$$We can write this as $$\sqrt{x^2+4}+|x-3|=5.$$If $x\le 3,$ then $$\sqrt{x^2+4}+(3-x)=5,$$so $\sqrt{x^2+4}=x+2.$ Squaring both sides, we get $$x^2+4=x^2+4x+4,$$which leads to $x=0.$ This solution corresponds to the origin.

If $x\ge 3,$ then $$\sqrt{x^2+4}+(x-3)=5,$$so $\sqrt{x^2+4}=8-x.$ Squaring both sides, we get $$x^2+4=64-16x+x^2,$$which leads to $x=\frac{15}{4}.$ Thus, the other $x$-intercept is $\boxed{\left(\frac{15}{4},0\right)}.$