Question

if x = `(sqrt(a + 1) + sqrt(a-1))/(sqrt(a + 1) - sqrt(a - 1))` using properties of proportion show that `x^2 - 2ax + 1 = 0`

Answer 1

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

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

By using componando and dividendo, we have

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

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

Squaring both sides we get

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

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

Again using componando and dividendo, we get

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

`=> (2x^2 + 2)/(4x) = (2a)/2 => (x^2 + 1)/(2x) = a/1`

`=> x^2 + 1 = 2ax`

`=> x^2 - 2ax + 1 = 0`

Answer 2

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

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

By componendo and dividendo

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

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

⇒ `(x + 1)^2/(x - 1)^2 = sqrt(a + 1)/sqrt(a - 1)` ...(by duplicate ratio)

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

Again by componendo and dividendo

`=> (x^2 + 2x + 1+ x^2 - 2x + 1)/(x^2 + 2x + 1 - x^2 + 2x - 1)`

= `(a + 1 + a - 1)/(a + 1- a - 1)`

⇒ `(x^2 + 1)/(2x) = a/1`

⇒ `x^2 + 1 = 2ax`

⇒ `x^2 - 2ax + 1 = 0`
Hence proved.