Question

If x = `(sqrt(2a + 1) + sqrt(2a - 1))/(sqrt(2a + 1) - sqrt(2a - 1))` prove that x² − 4ax + 1 = 0

Answer 1

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

Apply componendo and dividendo,

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

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

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

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

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

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

Apply componendo and dividendo again,

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

⇒ `(2x^2 + 2)/(4x) = (4a)/2`

⇒ `(x^2 + 1)/(2x) = 2a`

⇒ x² + 1 = 4ax

⇒ x² − 4ax + 1 = 0