Question

if x = root of (a+1) + root of (a-1) whole divided by root of (a+1) - root (a-1). show that x2-2ax+1 = 0

Answer 1

x=[(√a+1) + (√a-1 )]/[(√a+1)-(√a-1)]
On rationalising,
=[(√a+1)+(√a-1)][(√a+1)+(√a-1)]/[(√a+1)-(√a-1)][(√a+1)+(√a-1)]
=(a+1+a-1+2√a²-1)/a+1-a+1
=(2a+2√a²-1)/2=2(a+√a²-1)/2=a+√a²-1
x=a+√a²-1....(1)
x²-2ax+1=0
x²+(-2ax)+1=0
We know that,if ax²+bx+c=0,then x=(-b±√b²-4ac)/2a
x=-(-2a)±[√(-2a)²-4(1)(1)]/2    
=(2a±√4a²-4)/2=[2a±√4(a²-1)]/2
=(2a±2√a²-1)/2=a±√a²-1=a+√a²-1,a-√a²-1
∴x=a+√a²-1,a-√a²-1....(2)
∴(1)=(2)
Thus,verified.

Answer 2

Step-by-step explanation:

but I'm stuck there in the last step