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2014-05-13T09:39:22+05:30

given equation is 9(x^2 + 1/x^2) - 27(x + 1/x) + 8 = 0
let x + 1/x = t
9(t^2 - 2) - 27t + 8 = 0
9t^2 - 27t - 10 =0
(3t - 10)(3t + 1) = 0
t = 10/3 , -1/3
t = 10/3
x + 1/x = 10/3 
3x^2 - 10x + 3 = 0
(3x - 1)(x - 3) = 0
x = 3 , 1/3 
t = -1/3
x + 1/x = -1/3
3x^2 + x + 3 = 0.............................(1)
D = b^2 - 4ac
   = 1 - 36
   = -35                 (negative)
equation (1) is always positive ,hence no root 
root of the given equation is  x = 3, 1/3
 
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2014-05-13T09:57:28+05:30
The given equation is 9(x^2 + 1/x^2) - 27(x + 1/x) + 8 = 0
let x + \frac{1}{x}= t
9(t^2 - 2) - 27t + 8 = 0
9t^2 - 27t - 10 =0
(3t - 10)(3t + 1) = 0
t= \frac{10}{3} , - \frac{1}{3}
t =  \frac{10}{3}
x +  \frac{1}{x}  =  \frac{10}{3}
3x^2 - 10x + 3 = 0
(3x - 1)(x - 3) = 0
x = 3 ,  \frac{1}{3}
t = - \frac{1}{3}
x +  \frac{1}{x}  = - \frac{1}{3}
3x^2 + x + 3 = 0                                                                          (a)
D = b^2 - 4ac
   = 1 - 36
   = -35                 (negative)
Equation a is always positive ,therefore it has no root 
root of the given equation is  x = 3, 1/3

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