Question

If $$\frac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \frac{7}{3}$$, prove that $$x = 20$$.

Answer 1

Given: $$\frac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \frac{7}{3}$$

To prove: $$x = 20$$

Proof:

1. $$\frac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \frac{7}{3}$$ [Given]
2. $$\frac{(\sqrt{x + 5} + \sqrt{x - 16}) + (\sqrt{x + 5} - \sqrt{x - 16})}{(\sqrt{x + 5} + \sqrt{x - 16}) - (\sqrt{x + 5} - \sqrt{x - 16})} = \frac{7 + 3}{7 - 3}$$ [By Componendo & Dividendo]
3. or, $$\frac{2\sqrt{x + 5}}{2\sqrt{x - 16}} = \frac{10}{4}$$
4. or, $$\frac{\sqrt{x + 5}}{\sqrt{x - 16}} = \frac{5}{2}$$
5. or, $$\frac{x + 5}{x - 16} = \frac{25}{4}$$ [On squaring both sides]
6. or, $$\frac{(x + 5) + (x - 16)}{(x + 5) - (x - 16)} = \frac{25 + 4}{25 - 4}$$ [By Componendo & Dividendo]
7. or, $$\frac{2x - 11}{21} = \frac{29}{21}$$
8. or, $$2x - 11 = 29$$
9. or, $$2x = 40$$
10. or, $$x = 20$$

Hence proved.