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Solve 9("X"^2 + 1/"X"^2) -3("X" - 1/"X") - 20 = 0 - Algebra

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प्रश्न

Solve `9("x"^2 + 1/"x"^2) -3("x" - 1/"x") - 20 = 0`

योग

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`

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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2015-2016 (July)

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