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Simplify $\dfrac { \sqrt{ 3 } + \sqrt{ 2 } } { \sqrt{ 3 } - \sqrt{ 2 } }$.

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Answer:

$5 + 2 \sqrt{ 6 }$

Step by Step Explanation:

1. Let us multiply both the numerator and the denominator of the fraction by $(\sqrt{ 3 } + \sqrt{ 2 }).$
   $$\begin{align} & = \dfrac { \sqrt{ 3 } + \sqrt{ 2 } } { \sqrt{ 3 } - \sqrt{ 2 } } \times \dfrac { \sqrt{ 3 } + \sqrt{ 2 } } { \sqrt{ 3 } + \sqrt{ 2 } } \\ & = \dfrac { ( \sqrt { 3 } + \sqrt { 2 } )( \sqrt{ 3 } + \sqrt { 2 } ) } { (\sqrt{ 3 })^2 - (\sqrt { 2 })^2 } && [a^2 - b^2 = (a + b)(a - b)] \\ & = \dfrac { (\sqrt{ 3 } + \sqrt { 2 })^2 } { 3 - 2 } \\ & = (\sqrt{ 3 })^2 + (\sqrt{ 2 })^2 + 2 \times \sqrt{ 3 } \times \sqrt{ 2 } && [(a + b)^2 = a^2 + b^2 + 2ab] \\ & = 3 + 2 + 2\sqrt{ 6 } \\ & = 5 + 2 \sqrt{ 6 } \end{align}$$

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