Question

If $$\frac{\sqrt{x + 2} + \sqrt{x - 3}}{\sqrt{x + 2} - \sqrt{x - 3}} = 5$$, prove that $$x = 7$$.

Answer 1

Given: $$\frac{\sqrt{x + 2} + \sqrt{x - 3}}{\sqrt{x + 2} - \sqrt{x - 3}} = 5$$

To prove: $$x = 7$$

Proof:

1. $$\frac{\sqrt{x + 2} + \sqrt{x - 3}}{\sqrt{x + 2} - \sqrt{x - 3}} = \frac{5}{1}$$ [Given]
2. $$\frac{(\sqrt{x + 2} + \sqrt{x - 3}) + (\sqrt{x + 2} - \sqrt{x - 3})}{(\sqrt{x + 2} + \sqrt{x - 3}) - (\sqrt{x + 2} - \sqrt{x - 3})} = \frac{5 + 1}{5 - 1}$$ [By Componendo & Dividendo]
3. or, $$\frac{2\sqrt{x + 2}}{2\sqrt{x - 3}} = \frac{6}{4}$$
4. or, $$\frac{\sqrt{x + 2}}{\sqrt{x - 3}} = \frac{3}{2}$$
5. or, $$\frac{x + 2}{x - 3} = \frac{9}{4}$$ [On squaring both sides]
6. or, $$\frac{(x + 2) + (x - 3)}{(x + 2) - (x - 3)} = \frac{9 + 4}{9 - 4}$$ [By Componendo & Dividendo]
7. or, $$\frac{2x - 1}{5} = \frac{13}{5}$$
8. or, $$2x - 1 = 13$$
9. or, $$2x = 14$$
10. or, $$x = 7$$

Hence proved.