Question

If $$\frac{\sqrt{2a+3b} + \sqrt{2a-3b}}{\sqrt{2a+3b} - \sqrt{2a-3b}}$$ prove that $$3bx^2 - 4ax + 3b = 0$$.

Answer 1

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

To prove: $$3bx^2 - 4ax + 3b = 0$$

Proof:

1. $$\frac{x}{1} = \frac{\sqrt{2a + 3b} + \sqrt{2a - 3b}}{\sqrt{2a + 3b} - \sqrt{2a - 3b}}$$
2. or, $$\frac{x + 1}{x - 1} = \frac{(\sqrt{2a + 3b} + \sqrt{2a - 3b}) + (\sqrt{2a + 3b} - \sqrt{2a - 3b})}{(\sqrt{2a + 3b} + \sqrt{2a - 3b}) - (\sqrt{2a + 3b} - \sqrt{2a - 3b})}$$ [By componendo and dividendo]
3. or, $$\frac{x + 1}{x - 1} = \frac{2\sqrt{2a + 3b}}{2\sqrt{2a - 3b}}$$
4. or, $$\frac{x + 1}{x - 1} = \frac{\sqrt{2a + 3b}}{\sqrt{2a - 3b}}$$
5. or, $$\frac{(x + 1)^2}{(x - 1)^2} = \frac{2a + 3b}{2a - 3b}$$ [On squaring both sides]
6. or, $$\frac{(x + 1)^2 + (x - 1)^2}{(x + 1)^2 - (x - 1)^2} = \frac{(2a + 3b) + (2a - 3b)}{(2a + 3b) - (2a - 3b)}$$ [By componendo and dividendo]
7. or, $$\frac{2(x^2 + 1)}{4x} = \frac{4a}{6b}$$
8. or, $$\frac{x^2 + 1}{2x} = \frac{2a}{3b}$$
9. or, $$3b(x^2 + 1) = 4ax$$
10. or, $$3bx^2 + 3b = 4ax$$
11. or, $$3bx^2 - 4ax + 3b = 0$$

Hence proved.