Question

Using properties of proportion, find the value of x from the following:

`(a + sqrt(a^2 - 2ax))/(a - sqrt(a^2 - 2ax))` = b

Answer 1

`(a + sqrt(a^2 - 2ax))/(a - sqrt(a^2 - 2ax)) = b/1`

Applying componendo and dividendo,

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

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

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

Squaring both sides,

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

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

⇒ a² . (b − 1)² = (a² − 2ax) . (b + 1)²

⇒ a²(b² − 2b + 1) = a²(b2 + 2b + 1) − 2ax(b² + 2b + 1)

⇒ a²b² − 2a²b + a² = a²b² + 2a²b + a² − 2ax(b + 1)²

⇒ 2ax(b + 1)² = a²b² + 2a²b + a² − a²b² + 2a²b − a²

⇒ 2ax(b + 1)² = 4a²b

⇒ x = `(4a^2b)/(2a(b + 1)^2)`

⇒ x = `(2ab)/(b + 1)^2`