Question

if x=√a+1 + √a-1/√a+1 - √a-1 using componendo and dividendo....
state that : x2-2ax+1=0

(chapter: proportion)

Answer 1

x/1 = (√(a + 1) +√(a -1)/(√(a + 1) - √( a -1))

use componendo and dividendo
(x + 1)/(x - 1) = - √(a + 1)/√(a -1)

take square both sides ,

(x + 1)² /( x - 1)² = (a + 1)/(a - 1)

(a -1)(x+ 1)² = (x - 1)² ( a + 1)

a(x + 1)² -(x + 1)² = a(x - 1)² +(x - 1)²

a{(x + 1)² -(x - 1)² } = (x -1)² + (x + 1)²

a{ 4x } = 2(x ² + 1)

2ax = x² + 1

x² -2ax + 1 = 0

hence, proved