Question

$( \sqrt{x + 1 } + \sqrt{x - 1}) \div \sqrt{x + 1} - \sqrt{x - 1 = (4x - 1 \div 2$
componendo and dividendo​

Answer 1

Answer:

$\frac{ \sqrt{x + 1} + \sqrt{x - 1} }{ \sqrt{x + 1} - \sqrt{x - 1} } = (4x - 1 )\div 2$

If x/y=a/b

By using componendo dividendo,

$\frac{x + y}{x - y} = \frac{a + b}{a - b}$

So

$\frac{ \sqrt{x + 1} + \sqrt{x - 1} + \sqrt{x + 1} - \sqrt{x - 1} }{ \sqrt{x + 1} + \sqrt{x - 1} - \sqrt{x + 1 } + \sqrt{x - 1} } = \frac{4x - 1 + 2}{4x - 1 - 2}$

$\frac{ \sqrt{x + 1} }{ \sqrt{x - 1} } = \frac{4x + 1}{4x - 3}$

Squaring on both sides

$\frac{x + 1}{x - 1} = \frac{16 {x}^{2} + 1 + 8x}{16 {x}^{2} + 9 - 12x }$

$16 {x}^{3} + 9x - 12 {x}^{2} + 16 {x}^{2} + 9 - 12x = 16 {x}^{3} + x + 8 {x}^{2} - 16 {x}^{2} - 1 - 8x$

$12 {x}^{2} + 4x + 10 = 0$

$6 {x}^{2} + 2x + 5 = 0$