Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given expression is

[tex]\sf \: \dfrac{ \sqrt{2 - x} + \sqrt{2 + x} }{ \sqrt{2 - x} - \sqrt{2 + x} } = 3 \\ \\ [/tex]

can be rewritten as

[tex]\sf \: \dfrac{ \sqrt{2 - x} + \sqrt{2 + x} }{ \sqrt{2 - x} - \sqrt{2 + x} } = \frac{3}{1} \\ \\ [/tex]

We know,

If a : b :: c : d, then by Componendo and Dividendo,

[tex]\boxed{ \sf{ \:\dfrac{a + b}{a - b} = \dfrac{c + d}{c - d} \: }} \\ \\ [/tex]

So, on applying Componendo and Dividendo, we get

[tex]\sf \: \dfrac{ \sqrt{2 - x} + \sqrt{2 + x} + \sqrt{2 - x} - \sqrt{2 + x}}{ \sqrt{2 - x} + \sqrt{2 + x} - \sqrt{2 - x} + \sqrt{2 + x} } = \frac{3 + 1}{3 - 1} \\ \\ [/tex]

[tex]\sf \: \dfrac{ 2\sqrt{2 - x} }{2 \sqrt{2 + x} } = \frac{4}{2} \\ \\ [/tex]

[tex]\sf \: \dfrac{ \sqrt{2 - x} }{\sqrt{2 + x} } = \frac{2}{1} \\ \\ [/tex]

On squaring both sides, we get

[tex]\sf \: \dfrac{2 - x}{2 + x} = \dfrac{4}{1} \\ \\ [/tex]

[tex]\sf \: 8 + 4x = 2 - x \\ \\ [/tex]

[tex]\sf \: x + 4x = 2 - 8 \\ \\ [/tex]

[tex]\sf \: 5x = - 6 \\ \\ [/tex]

[tex]\bf\implies \:x = \: - \: \dfrac{6}{5} \\ \\ [/tex]

Hence,

[tex]\bf\implies \:The \: value \: of \: x = \: - \: \dfrac{6}{5} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

If a : b :: c : d, then

1. Alternendo

[tex]\sf \: \dfrac{a}{c} = \dfrac{b}{d} \\ \\ \\ [/tex]

2. Invertendo

[tex]\sf \: \dfrac{b}{a} = \dfrac{d}{c} \\ \\ \\ [/tex]

3. Componendo

[tex]\sf \: \dfrac{a + b}{b} = \dfrac{c + d}{d} \\ \\ \\ [/tex]

4. Dividendo

[tex]\sf \: \dfrac{a - b}{b} = \dfrac{c - d}{d} \\ \\ \\ [/tex]