Q.If \(x=\cfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}\), then show that \(bx^2 - ax + b = 0\).

\(x = \cfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}\)
Or, \(\cfrac{x}{1} = \cfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}\)
Or, \(\cfrac{x+1}{x-1} = \cfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}\)

\(\cfrac{+\sqrt{a+2b}-\sqrt{a-2b}}{-\sqrt{a+2b}+\sqrt{a-2b}}\)

[Using the addition-subtraction process]

Or, \(\cfrac{x+1}{x-1} = \cfrac{2\sqrt{a+2b}}{2\sqrt{a-2b}}\)
Or, \(\cfrac{x+1}{x-1} = \cfrac{\cancel2\sqrt{a+2b}}{\cancel2\sqrt{a-2b}}\)
Or, \(\left(\cfrac{x+1}{x-1}\right)^2 = \left(\cfrac{\sqrt{a+2b}}{\sqrt{a-2b}}\right)^2\)
Or, \(\cfrac{x^2+2x+1}{x^2-2x+1} = \cfrac{a+2b}{a-2b}\)
Or, \(\cfrac{(x^2+2x+1)+(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)}\)

\(=\cfrac{(a+2b)+(a-2b)}{(a+2b)-(a-2b)}\)

[Applying the addition-subtraction process again]

Or, \(\cfrac{x^2+2x+1+x^2-2x+1}{x^2+2x+1-x^2+2x-1}\)

\(=\cfrac{a+2b+a-2b}{a+2b-a+2b}\)

Or, \(\cfrac{2x^2+2}{4x} = \cfrac{2a}{4b}\)
Or, \(\cfrac{x^2+1}{2x} = \cfrac{a}{2b}\)
Or, \(\cfrac{x^2+1}{x} = \cfrac{a}{b}\)
Or, \(bx^2 + b = ax\)
Or, \(bx^2 - ax + b = 0\) [Proved]
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