Question

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

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

Answer 1

We have,

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

Applying componendo and dividendo,

⇒ `[(sqrt(1 + x) + sqrt(1 - x)) + (sqrt(1 + x) - sqrt(1 - x))]/[(sqrt(1 + x) + sqrt(1 - x)) - (sqrt(1 + x) - sqrt(1 - x))] = (a + b)/(a - b)`

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

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

Squaring both sides,

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

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

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

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

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

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

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

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

⇒ `x = (4ab)/(2a^2 + 2b^2)`

⇒ `x = (4ab)/(2(a^2 + b^2))`

⇒ `x = (2ab)/(a^2 + b^2)`