Question

If `x = (sqrt(a + 3b) + sqrt(a - 3b))/(sqrt(a + 3b) - sqrt(a - 3b))`, prove that: 3bx² – 2ax + 3b = 0.

Answer 1

`x = (sqrt(a + 3b) + sqrt(a - 3b))/(sqrt(a + 3b) - sqrt(a - 3b))`

Applying componendo and dividendo, we get,

`(x + 1)/(x - 1) = (sqrt(a + 3b) + sqrt(a - 3b) + sqrt(a + 3b) - sqrt(a - 3b))/(sqrt(a + 3b) + sqrt(a - 3b) - sqrt(a + 3b) + sqrt(a - 3b))`

`(x + 1)/(x - 1)= (2sqrt(a + 3b))/(2sqrt(a - 3b))`

Squaring both sides,

`(x^2 + 2x + 1)/(x^2 - 2x + 1) = (a + 3b)/(a - 3b)`

Again applying componendo and dividendo,

`(x^2 + 2x + 1 + x^2 - 2x + 1)/(x^2 + 2x + 1 - x^2 + 2x - 1) = (a + 3b + a - 3b)/(a + 3b - a + 3b)`

`(2(x^2 + 1))/(2(2x)) = (2(a))/(2(3b))`

3b(x² + 1) = 2ax

3bx² + 3b = 2ax

3bx² – 2ax + 3b = 0