If x=(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b)) … then prove that bx2-ax+b=0

x =(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b))

use Componendo and dividendo

(x+1)/(x-1) = √(a+2b)/√(a-2b)

square both sides

(x^2 + 2 x + 1)/(x^2-2x + 1) = (a+2b)/(a-2b)

again use Componendo and dividendo

(x^2 + 1)/(2x) = (a)/(2b)

or bx^2 +b = ax
or bx^2 - ax + b = 0

Note: other method can be used but it is simpler when we have form (a+b)/(a-b)

Source(s):

http://en.wikipedia.org/wiki/Componendo_…