If x = (√(2a + 1) + √(2a - 1)) / (√(2a + 1) - √(2a - 1)), prove that x² - 4ax + 1 = 0

Let p = √(2a + 1) and q = √(2a − 1).

Given:

x = (p + q)/(p − q)

Therefore: 1/x = (p − q)/(p + q)

Add the two expressions:

x + 1/x = [(p + q)² + (p − q)²] / [(p + q)(p − q)]

= [2p² + 2q²] / (p² − q²)

Now: p² + q² = (2a + 1) + (2a − 1) = 4a

and

p² − q² = (2a + 1) − (2a − 1) = 2

Hence: x + 1/x = 2(4a)/2 = 4a

Multiplying by x:

x² + 1 = 4ax

Therefore:

x² − 4ax + 1 = 0

Hence proved.

Let p = √(2a + 1) and q = √(2a − 1). Adding x and 1/x makes the radical expressions combine through p² + q² and p² − q², which simplify to 4a and 2 respectively.