Question

If x = `(sqrt(2a + 3b) + sqrt(2a - 3b))/(sqrt(2a + 3b) - sqrt(2a - 3b))` then show that 3bx² − 4ax + 3b = 0.

Answer 1

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

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

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

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

Square both sides,

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

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

Apply Componendo and Dividendo again,

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

⇒ `(2x^2 + 2)/(4x) = (4a)/(6b)`

⇒ `(2(x^2 + 1))/(4x) = (4a)/(6b)`

⇒ `(x^2 + 1)/(2x) = (2a)/(3b)`

⇒ 3b(x² + 1) = 2a(2x)

⇒ 3bx² + 3b = 4ax

Rearranging the terms to form the quadratic equation:

⇒ 3bx² − 4ax + 3b = 0