Question 877417
30(x^2+1/x^2)-77(x-1/x)-12=0

{{{30(x^2+1/x^2)-77(x+1/x)-12}}}{{{""=""}}}{{{"0"}}}
<pre>
Let {{{u}}}{{{""=""}}}{{{x+1/x}}}

Then

    {{{u^2}}}{{{""=""}}}{{{(x+1/x)^2}}}{{{""=""}}}{{{x^2+2+1/x^2}}}

    {{{u^2}}}{{{""=""}}}{{{x^2+1/x^2+2}}}

    {{{u^2-2}}}{{{""=""}}}{{{x^2+1/x}}}

So substitute {{{u^2-2}}} for {{{x^2+1/x}}}
and {{{u}}} for {{{x+1/x}}}

{{{30(u^2-2)-77u-12}}}{{{""=""}}}{{{"0"}}}

{{{30u^2-60-77u-12}}}{{{""=""}}}{{{"0"}}}

{{{30u^2-77u-72}}}{{{""=""}}}{{{"0"}}}

Use the quadratic formula, get

((u=(77 +- sqrt(14569))/60}}}


{{{x+1/x}}}{{{""=""}}}{{{(77 +- sqrt(14569))/60}}}

Solve that and get

{{{x = 1/120 (77-sqrt(14569) +- i*sqrt(2(77sqrt(14569)-3049)))}}}

It's too complicated.  I believe you may have copied something wrong.

This has four complicated complex roots and I don't believe your
teacher would have given you such a laborious problem.
Edwin</pre>