Question

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

Answer 1

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

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

Proof:

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

Hence proved.