More Square Roots (Medium)

$\dfrac{\sqrt{6} - 2\sqrt{3}}{\sqrt{6} + 2\sqrt{3}} {\color{#3D99F6}{\times \dfrac{\sqrt{6} - 2\sqrt{3}}{\sqrt{6} - 2\sqrt{3}} }} \implies \dfrac{\big(\sqrt{6} - 2\sqrt{3}\big)^2}{(\sqrt{6})^2 - (2\sqrt{3})^2} = \dfrac{6 - 4\sqrt{8} + 12}{6 - 12} = \dfrac{\cancel{-6} \big( -3 + 2\sqrt{2} \big)}{\cancel{-6}} = \boxed{ -3 + 2\sqrt{2} }$

$\frac { \sqrt { 6 } -2\sqrt { 3 } }{ \sqrt { 6 } +2\sqrt { 3 } }$ Then to rationalize the denominator you multiply both numerator and denominator by the conjugate. $\sqrt { 6 } -2\sqrt { 3 }$ .

$\frac { \sqrt { 6 } -2\sqrt { 3 } }{ \sqrt { 6 } +2\sqrt { 3 } } *\frac { \sqrt { 6 } -2\sqrt { 3 } }{ \sqrt { 6 } -2\sqrt { 3 } } =\frac { 6-12\sqrt { 2 } +12 }{ -6 } =-3+2\sqrt { 2 }$