#### 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`

#### Solution

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

By applying Componendo-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) = (2sqrt(a + 1))/(2sqrt(a - 1))`

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

Squaring both the sides of the equation we have

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

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

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

`=> 4ax = 2x^2 + 2`

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

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

Is there an error in this question or solution?

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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` Concept: Componendo and Dividendo Properties.

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