Question

Using Componendo and Dividendo solve for x:

${\displaystyle \frac{\sqrt{2x+2}+\sqrt{2x-1}}{\sqrt{2x+2}-\sqrt{2x-1}}}$ = 3

Answer 1

${\displaystyle \frac{\sqrt{2x+2}+\sqrt{2x-1}}{\sqrt{2x+2}-\sqrt{2x-1}}=\frac{3}{1}}$

Using componendo and dividendo,

${\displaystyle \frac{(\sqrt{2x+2}+\sqrt{2x-1})+(\sqrt{2x+2}-\sqrt{2x-1})}{(\sqrt{2x+2}+\sqrt{2x-1})-(\sqrt{2x+2}-\sqrt{2x-1})}=\frac{3+1}{3-1}}$

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

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

On squaring

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

2x + 2 = 8x – 4

2x – 8x = – 2 – 4

– 6x =  – 6

x = 1