Answer: A
Let \(x = \cfrac{\sqrt{a + 2b} + \sqrt{a - 2b}}{\sqrt{a + 2b} - \sqrt{a - 2b}}\) i.e., \(\cfrac{x}{1} = \cfrac{\sqrt{a + 2b} + \sqrt{a - 2b}}{\sqrt{a + 2b} - \sqrt{a - 2b}}\) i.e., \(\cfrac{x + 1}{x - 1} = \cfrac{\sqrt{a + 2b} + \sqrt{a - 2b}}{\sqrt{a + 2b} - \sqrt{a - 2b}} \cdot \cfrac{\sqrt{a + 2b} - \sqrt{a - 2b}}{-\sqrt{a + 2b} + \sqrt{a - 2b}}\) i.e., \(\cfrac{x + 1}{x - 1} = \cfrac{2\sqrt{a + 2b}}{2\sqrt{a - 2b}}\) i.e., \(\cfrac{x + 1}{x - 1} = \cfrac{\sqrt{a + 2b}}{\sqrt{a - 2b}}\) i.e., \(\left(\cfrac{x + 1}{x - 1}\right)^2 = \left(\cfrac{\sqrt{a + 2b}}{\sqrt{a - 2b}}\right)^2\) i.e., \(\cfrac{x^2 + 2x + 1}{x^2 - 2x + 1} = \cfrac{a + 2b}{a - 2b}\) Now, \(\cfrac{(x^2 + 2x + 1) + (x^2 - 2x + 1)}{(x^2 + 2x + 1) - (x^2 - 2x + 1)} = \cfrac{(a + 2b) + (a - 2b)}{(a + 2b) - (a - 2b)}\) i.e., \(\cfrac{2x^2 + 2}{4x} = \cfrac{2a}{4b}\) i.e., \(\cfrac{x^2 + 1}{2x} = \cfrac{a}{2b}\) i.e., \(\cfrac{x^2 + 1}{x} = \cfrac{a}{b}\) i.e., \(bx^2 + b = ax\) i.e., \(bx^2 - ax + b = 0\)
Let \(x = \cfrac{\sqrt{a + 2b} + \sqrt{a - 2b}}{\sqrt{a + 2b} - \sqrt{a - 2b}}\) i.e., \(\cfrac{x}{1} = \cfrac{\sqrt{a + 2b} + \sqrt{a - 2b}}{\sqrt{a + 2b} - \sqrt{a - 2b}}\) i.e., \(\cfrac{x + 1}{x - 1} = \cfrac{\sqrt{a + 2b} + \sqrt{a - 2b}}{\sqrt{a + 2b} - \sqrt{a - 2b}} \cdot \cfrac{\sqrt{a + 2b} - \sqrt{a - 2b}}{-\sqrt{a + 2b} + \sqrt{a - 2b}}\) i.e., \(\cfrac{x + 1}{x - 1} = \cfrac{2\sqrt{a + 2b}}{2\sqrt{a - 2b}}\) i.e., \(\cfrac{x + 1}{x - 1} = \cfrac{\sqrt{a + 2b}}{\sqrt{a - 2b}}\) i.e., \(\left(\cfrac{x + 1}{x - 1}\right)^2 = \left(\cfrac{\sqrt{a + 2b}}{\sqrt{a - 2b}}\right)^2\) i.e., \(\cfrac{x^2 + 2x + 1}{x^2 - 2x + 1} = \cfrac{a + 2b}{a - 2b}\) Now, \(\cfrac{(x^2 + 2x + 1) + (x^2 - 2x + 1)}{(x^2 + 2x + 1) - (x^2 - 2x + 1)} = \cfrac{(a + 2b) + (a - 2b)}{(a + 2b) - (a - 2b)}\) i.e., \(\cfrac{2x^2 + 2}{4x} = \cfrac{2a}{4b}\) i.e., \(\cfrac{x^2 + 1}{2x} = \cfrac{a}{2b}\) i.e., \(\cfrac{x^2 + 1}{x} = \cfrac{a}{b}\) i.e., \(bx^2 + b = ax\) i.e., \(bx^2 - ax + b = 0\)