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प्रश्न
If `x = (sqrt(a + 3b) + sqrt(a - 3b))/(sqrt(a + 3b) - sqrt(a - 3b))`, prove that: 3bx2 – 2ax + 3b = 0.
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उत्तर
`x = (sqrt(a + 3b) + sqrt(a - 3b))/(sqrt(a + 3b) - sqrt(a - 3b))`
Applying componendo and dividendo, we get,
`(x + 1)/(x - 1) = (sqrt(a + 3b) + sqrt(a - 3b) + sqrt(a + 3b) - sqrt(a - 3b))/(sqrt(a + 3b) + sqrt(a - 3b) - sqrt(a + 3b) + sqrt(a - 3b))`
`(x + 1)/(x - 1)= (2sqrt(a + 3b))/(2sqrt(a - 3b))`
Squaring both sides,
`(x^2 + 2x + 1)/(x^2 - 2x + 1) = (a + 3b)/(a - 3b)`
Again applying componendo and dividendo,
`(x^2 + 2x + 1 + x^2 - 2x + 1)/(x^2 + 2x + 1 - x^2 + 2x - 1) = (a + 3b + a - 3b)/(a + 3b - a + 3b)`
`(2(x^2 + 1))/(2(2x)) = (2(a))/(2(3b))`
3b(x2 + 1) = 2ax
3bx2 + 3b = 2ax
3bx2 – 2ax + 3b = 0
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