Question 1187475
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Fun problem...!<br>
Here is the key:<br>
{{{x^2+1/x^2 = (x^2+2+1/x^2)-2 = (x+1/x)^2-2}}}<br>
Use that to rewrite the first expression in parentheses and you end up with a quadratic equation with "x+1/x" as the "variable".<br>
Solve that equation for x+1/x and you end up solving two quadratic equations that each have two real solutions, giving you four solutions to the given equation.<br>
{{{12(x^2+1/x^2)-56(x+1/x)+89=0}}}
{{{12(x^2+2+1/x^2)-12(2)-56(x+1/x)+89=0}}}
{{{12(x+1/x)^2-24-56(x+1/x)+89=0}}}
{{{12(x+1/x)^2-56(x+1/x)+65=0}}}<br>
That's a quadratic equation in the "variable" x+1/x -- solve it by factoring.<br>
{{{(2(x+1/x)-5)(6(x+1/x)-13)=0}}}<br>
{{{x+1/x=5/2}}} or {{{x+1/x=13/6}}}<br>
Solve each of those equations....<br>
{{{x+1/x=5/2}}}
{{{2x+2/x=5}}}
{{{2x-5+2/x=0}}}
{{{2x^2-5x+2=0}}}
{{{(2x-1)(x-2)=0}}}
{{{x=1/2}}} or {{{x=2}}}<br>
There are two solutions; note that they are reciprocals of each other.  That makes sense, since the equation can be written as a quadratic with "variable" (x+1/x) -- if x=2 then 1/x=1/2; and if x=1/2 then 1/x=2.<br>
{{{x+1/x=13/6}}}
{{{6x+6/x=13}}}
{{{6x-13+6/x=0}}}
{{{6x^2-13x+6=0}}}
{{{(2x-3)(3x-2)=0}}}
{{{x=3/2}}} or {{{x=2/3}}}<br>
There are two more solutions which are also reciprocals of each other.<br>
ANSWER: There are 4 solutions: x=2 or x=1/2; and x=3/2 and x=2/3<br>
Here is a graph of the given equation, showing the four zeros at 1/2, 2/3, 3/2, and 2.<br>
{{{graph(600,200,-1,4,-1,1,12(x^2+1/x^2)-56(x+1/x)+89)}}}<br>