By the geometric mean theorem,
[tex]\frac{y}{4}=\frac{12}{y} \\ \\ y^{2}=48 \\ \\ \boxed{y=4\sqrt{3}}[/tex]
So, by the Pythagorean theorem,
[tex]12^{2}+(4\sqrt{3})^{2}=x^{2}\\\\144+48=x^{2}\\\\192=x^{2}\\\\\boxed{x=8\sqrt{3}}[/tex]
