If $$x = \dfrac{{\sqrt {2a + 1}  + \sqrt {2a - 1} }}{{\sqrt {2a + 1}  - \sqrt {2a - 1} }}$$, then show that $${x^2} - 4ax + 1 = 0$$

Solution

$$\begin{array}{l}x = \cfrac{{\sqrt {2a + 1}  + \sqrt {2a - 1} }}{{\sqrt {2a + 1}  - \sqrt {2a - 1} }}\\ = \cfrac{{\left( {\sqrt {2a + 1}  + \sqrt {2a - 1} } \right)}}{{\left( {\sqrt {2a + 1}  - \sqrt {2a - 1} } \right)}}.\cfrac{{\left( {\sqrt {2a + 1}  + \sqrt {2a - 1} } \right)}}{{\left( {\sqrt {2a + 1}  + \sqrt {2a - 1} } \right)}}\\ = \cfrac{{{{\left( {\sqrt {2a + 1}  + \sqrt {2a - 1} } \right)}^2}}}{{2a + 1 - 2a + 1}}\\ = \cfrac{{2a + 1 + 2a - 1 + 2\sqrt {\left( {2a + 1} \right)\left( {2a - 1} \right)} }}{2}\\ \Rightarrow 2x = 4a + 2\sqrt {{{\left( {2a} \right)}^2} - 1} \\ \Rightarrow x = 2a + \sqrt {4{a^2} - 1} \\ \Rightarrow x - 2a = \sqrt {4{a^2} - 1} \\ \Rightarrow {\left( {x - 2a} \right)^2} = 4{a^2} - 1\\ \Rightarrow {x^2} + 4{a^2} - 4ax = 4{a^2} - 1\\ \Rightarrow {x^2} - 4ax + 1 = 0\end{array}$$