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If x=√2a+3b+√...

Question

If $x = \frac{\sqrt{2 a + 3 b} + \sqrt{2 a - 3 b}}{\sqrt{2 a + 3 b} - \sqrt{2 a - 3 b}}$ , then show that $3 b x^{2} - 4 a x + 3 b = 0$ .

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Solution

Proof:
If $\frac{a}{b} = \frac{c}{d}$ , then the property of componendo and dividendo states that $\frac{a + b}{a - b} = \frac{c + d}{c - d}$ .
The given expression can be rewritten as,
$\frac{x}{1} = \frac{\sqrt{2 a + 3 b} + \sqrt{2 a - 3 b}}{\sqrt{2 a + 3 b} - \sqrt{2 a - 3 b}} \Rightarrow \frac{x + 1}{x - 1} = \frac{\sqrt{2 a + 3 b} + \sqrt{2 a - 3 b} + \sqrt{2 a + 3 b} - \sqrt{2 a - 3 b}}{\sqrt{2 a + 3 b} + \sqrt{2 a - 3 b} - \sqrt{2 a + 3 b} + \sqrt{2 a - 3 b}} (B y c o m p o n e n d o a n d d i v i d e n d o) \Rightarrow \frac{x + 1}{x - 1} = \frac{2 \sqrt{2 a + 3 b}}{2 \sqrt{2 a - 3 b}} \Rightarrow \frac{x + 1}{x - 1} = \frac{\sqrt{2 a + 3 b}}{\sqrt{2 a - 3 b}} \Rightarrow \frac{{(x + 1)}^{2}}{{(x - 1)}^{2}} = \frac{2 a + 3 b}{2 a - 3 b} (S q u a r i n g b o t h s i d e s) \Rightarrow \frac{{(x + 1)}^{2} + {(x - 1)}^{2}}{{(x + 1)}^{2} - {(x - 1)}^{2}} = \frac{2 a + 3 b + 2 a - 3 b}{2 a + 3 b - (2 a - 3 b)} (B y c o m p o n e n d o a n d d i v i d e n d o) \Rightarrow \frac{2 (x^{2} + 1)}{4 x} = \frac{4 a}{6 b} (U \sin g t h e f o r m u l a e {(a - b)}^{2} = a^{2} - 2 a b + b^{2} a n d {(a + b)}^{2} = a^{2} + 2 a b + b^{2}) \Rightarrow \frac{x^{2} + 1}{2 x} = \frac{2 a}{3 b} \Rightarrow 3 b (x^{2} + 1) = 2 a \cdot 2 x (C r o s s m u l t i p l y i n g b o t h s i d e s) \Rightarrow 3 b x^{2} + 3 b = 4 a x \Rightarrow 3 b x^{2} - 4 a x + 3 b = 0 (S u b t r a c t i n g 4 a x f r o m b o t h s i d e s)$ ,
Hence proved.

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x = \frac{\sqrt{2 a + 3 b} + \sqrt{2 a - 3 b}}{\sqrt{2 a + 3 b} - \sqrt{2 a - 3 b}}

, the prove that

3 b x^{2} - 4 a x + 3 b = 0

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