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प्रश्न
Solve:
`9(x^2 + 1/x^2) - 9(x + 1/x) - 52 = 0`
उत्तर
Given equation be,
`9(x^2 + 1/x^2) - 9(x + 1/x) - 52 = 0`
Put `x + 1/x = y`
Then `x^2 + 1/x^2 + 2 = y^2` ...[On squaring]
`\implies x^2 + 1/x^2 = y^2 - 2`
Thus given equation
∴ `9(x^2 + 1/x^2) - 9(x + 1/x) - 52 = 0`
`\implies` 9(y2 – 2) – 9y – 52 = 0
`\implies` 9y2 – 18 – 9y – 52 = 0
`\implies` 9y2 – 9y – 70 = 0
`\implies` 9y2 – 30y + 21y – 70 = 0
`\implies` 3y(3y – 10) + 7(3y – 10) = 0
`\implies` (3y – 10)(3y + 7) = 0
Either 3y – 10 = 0, then y = `10/3`
Or 3y + 7 = 0 then y = `(-7)/3`
(i) When y = `10/3`, then `x + 1/x = 10/3`
`\implies` 3x2 + 3 = 10x
`\implies` 3x2 – 10x + 3 = 0
`\implies` 3x2 – 9x – x + 3 = 0
`\implies` 3x(x – 3) – 1(x – 3) = 0
`\implies` (x – 3)(3x – 1) = 0
Either x – 3 = 0, then x = 3
Or 3x – 1 = 0, then x = `1/3`
(ii) When `y = (-7)/3`, then `x + 1/x = (-7)/3`
`\implies` 3x2 + 3 = –7x
`\implies` 3x2 + 7x + 3 = 0
Here a = 3, b = 7, c = 3
∴ D = b2 – 4ac
= (7)2 – 4 × 3 × 3
= 49 – 36
= 13
∴ `x = (-b xx sqrt(D))/(2a)`
= `(-7 +- sqrt(13))/(2 xx 3)`
= `(-7 +- sqrt(13))/6`
∴ x = `3, 1/3, (-7 +- sqrt(13))/6`
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