From the Pythagorean theorem, we have
[tex](2x)^2 + (3x)^2 = (\sqrt{52})^2[/tex]
Simplifying both sides gives
[tex]4x^2 + 9x^2 = 52 \\\\ 13x^2 = 52 \\\\ x^2 = \dfrac{52}{13} = 4 \\\\ x = \pm\sqrt4 = \pm2[/tex]
But we're talking about lengths, which can't be negative, so x = 2.
