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प्रश्न
If `x = (sqrt(a + 1) + sqrt(a - 1))/(sqrt(a + 1) - sqrt(a - 1))`, using properties of proportion show that: x2 – 2ax + 1 = 0.
उत्तर
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)
`=>` (x2 + 2x + 1)(a – 1) = (x2 – 2x + 1)(a + 1)
`=>` a(x2 + 2x + 1) – (x2 + 2x + 1) = a(x2 – 2x + 1) + (x2 – 2x + 1)
`=>` 4ax = 2x2 + 2
`=>` 2ax = x2 + 1
`=>` x2 – 2ax + 1 = 0
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