Question

If `x = (sqrt(a + 1) + sqrt(a - 1))/(sqrt(a + 1) - sqrt(a - 1))`, using properties of proportion show that: x² – 2ax + 1 = 0.

Answer 1

Given that `x = (sqrt(a + 1) + sqrt(a - 1))/(sqrt(a + 1) - sqrt(a- 1))`

By applying componendo-dividendo,

`=> (x + 1)/(x - 1) = ((sqrt(a + 1) + sqrt(a - 1)) + (sqrt(a + 1) + sqrt(a - 1)))/((sqrt(a + 1) + sqrt(a - 1)) - (sqrt(a + 1) - sqrt(a - 1)))`

`=> (x + 1)/(x - 1) = (2sqrt(a + 1))/(2sqrt(a - 1))`

`=> (x + 1)/(x - 1)  = sqrt(a + 1)/sqrt(a -1 )`

Squaring both the sides of the equation, we have,

`=> ((x + 1)/(x - 1))^2 = (a + 1)/(a - 1)`

`=>` (x + 1)²(a – 1) = (x – 1)²(a + 1)

`=>` (x² + 2x + 1)(a – 1) = (x² – 2x + 1)(a + 1)

`=>` a(x² + 2x + 1) – (x² + 2x + 1) = a(x² – 2x + 1) + (x² – 2x + 1)

`=>` 4ax = 2x² + 2

`=>` 2ax = x² + 1

`=>` x² – 2ax + 1 = 0