[tex]\sqrt{a}=b\iff b^2=a\qquad for\ a\geq0\ and\ b\geq0\\\\\sqrt8-\sqrt9=\sqrt{4\cdot2}-3=\sqrt4\cdot\sqrt2-3=\boxed{2\sqrt2-3}\\\\\sqrt9=3\ because\ 3^2=9\\\\\sqrt4=2\ because\ 2^2=4\\\\Used\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\Answer:\ \boxed{\sqrt8-\sqrt9=2\sqrt2-3}[/tex]
