Question 1046945
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Kindly help me solving it i tried my best but I couldn't 
Solving equation (x-1/x)^2-6 (x+1/x)+12=0 we get roots as?
I tried to simplify it and it reduced to
x^4-6x^3+14x^2-6x+1=0
But i can't go further
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<pre>
It is elementary . . . 

0.  {{{(x-1/x)^2-6 (x+1/x)+12}}} = 0.    (1)


1.  {{{(x-1/x)^2}}} = {{{x^2 -2}}} + {{{1/x^2}}} = {{{x^2 + 2}}} + {{{1/x^2}}} -4 = {{{(x+1/x)^2}}} - 4.


2.  Therefore, you can write the original equation in an equivalent form

    {{{(x+1/x)^2 - 4 - 6(x + 1/x) + 12}}} = 0,  or, which is the same

    {{{(x+1/x)^2 - 6(x + 1/x) + 8}}} = 0.     (2)


3.  Next, introduce new variable y = {{{x + 1/x}}}. Then the equation (2) takes the form

    {{{y^2 -6y +8}}} = 0,

    and you can solve it by factoring

    ((y-4)*(y-2) = 0.

    The roots are y=4  and  y=2.


4.  As the last, final step, you should solve these two separate equations     

       {{{x + 1/x}}} = 6,

    and

       {{{x + 1/x}}} = 2.
</pre>

I believe you can complete the assignment from this point on your own.