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प्रश्न
Solve `9("x"^2 + 1/"x"^2) -3("x" - 1/"x") - 20 = 0`
उत्तर
`9[("x" - 1/"x")^2 + 2] -3 ("x" - 1/"x") - 20 = 0`
Let x - `1/"x" = "m"`
`9("m"^2 + 2) -3"m" -20 = 0`
`9"m"^2 + 18 - 3"m" - 20 = 0`
`9"m"^2 - 3"m" - 2 = 0`
9m2 - 6m + 3m - 2 = 0
3m (3m - 2) + 1(3m - 2) = 0
(3m - 2) (3m + 1) = 0
3m - 2 = 0 Or 3m + 1 = 0
3m = 2 Or 3m = -1
m = `2/3` Or m = `(-1)/3`
Resubstituting m = x - `1/x`
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m = `2/3` x - `1/x = 2/3` Or `(x^2 - 1)/x = 2/3` 3x2 - 3 = 2x 3x2 - 2x - 3 = 0 Comparing with ax2 + bx + c = 0 a = 3, b = -2, c = -3 b2 - 4ac = (-2)2 - 4 (3) (-3) = 4 + 36 = 40 By formula method x = `-b ± sqrt(b^2 - 4ac)/"2a"` = `(2 ± sqrt40)/"2(3)"` = `(2 ± 2sqrt10)/6` = `(2 (1 ± sqrt10))/6` x = `(1 ± sqrt10)/3` |
m = `(-1)/3` x - `1/x = (-1)/3` `("x"^2 - 1)/"x" = (-1)/3` 3x2 - 3 = -x Comparing with ax2 + bx + c = 0 a = 3, b = 1, c = -3 b2 - 4ac = 12 - 4 (3) (-3) = 1+ 36 = 37 By formula method x = `-b ± sqrt(b^2 - 4ac)/"2a"` = `(-1 ± sqrt37)/"2(3)"` x = `(-1 ± sqrt37)/6` |
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