x= sqrt(sqrt(5)+1)/(sqrt(5)-1)

prove that:

x^2-x-1=0

Let us try and simplify x:

x= sqrt(sqrt(5)+1)/sqrt(sqrt(5)-1)

= sqrt[sqrt(5)+1/sqrt(5)-1}

Multiply by sqrt(5)+1

= sqrt [(sqrt(5)+1)^2 / 4

= sqrt(5)+1 /2........(1)

Now let us find the function's root:

x^2-x-1=0

x1 = 1+sqrt(1-4(-1)= (1+ sqrt(5))/2........(2)

x2= (1-sqrt(5))/2

We notice that x is a root for the function x^2-x-1.

If x = sqrt((sqrt(5)+1)/(sqrt(5)-1)), then prove that x^2-x-1=0.

Solution:

Given x = sqrt((sqrt(5)+1)/(sqrt(5)-1))

= sqrt{ [5^(1/2) +1]/(5^1/2)-1)]}. Rationalising inside the bracket,

sqrt{ (5^(1/2)+1)(5^(1/2)+1/ ((5^1/2)^2-1)}

= sqrt{5^(1/2)+1)^2/4}. So,

x =(sqrt5)+1)/2 must be a root of x^2-x -1 = 0.........(1)

We start from x^2-x-1 =o and find the solution for x and see whether one of the roots of this quadratic equation, x^2-x-1 = 0 is th value of x as at eq(1):

x^2-x+(1/2)^2 - (1/2)^2 - 1 =. We added and subtracted (1/2) ^2 so that x^2-x+(1/2)^2 = (x-1/2)^2 is a perfect square.

Therefore, (x-1/2)^2 - (1/4+1) = 0 . Or

(x-1/2)^2 = 5/4. Taking square root on both sides, we get:

x1-1/2 = sqrt(5/4) . Or x2 -1/2= sqrt(5/4). Or

x1 = (1/2)+sqrt(5/4) = [(sqrt5)+1]/2 ia root of the equation which is the samevalue of x as at eq(1).