Question 894668
Maybe the results you are getting are the same as one of the options, but expressed in an equivalent form.
There are many ways to solve the equation.
I would start by multiplying both sides of the equal sign times {{{x^2}}} ,
and then I would solve the resulting quadratic equation:
{{{1+1/x^2=3/x}}}
{{{x^2(1+1/x^2)=x^2(3/x)}}}
{{{x^2+1=3x}}}
{{{x^2-3x+1=0}}}
One way to solve that quadratic equation is by completing the square>
Since {{{x^2-3x}}} is part of the square {{{x^2-3x+9/4=(x-3/2)^2}}} ,
we can complete the square by adding {{{9/4}}} .
of course, we add it to both sides of the equation, to get an equivalent equation:
{{{x^2-3x+1=0}}}
{{{x^2-3x=-1}}}
{{{x^2-3x+9/4=-1+9/4}}}
{{{(x-3/2)^2=5/4}}}
{{{(x-3/2)^2=(sqrt(5)/2)^2}}}
{{{system(x-3/2=sqrt(5)/2,"or",x-3/2=-sqrt(5)/2)}}} ---> {{{system(highlight(x=3/2+sqrt(5)/2),"or",highlight(x=3/2-sqrt(5)/2))}}
 
You may prefer to solve {{{x^2-3x+1=0}}} by using the quadratic formula,
which says that the solutions to {{{ax^2+b3x+c=0}}} are given by {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} .
In the case of {{{x^2-3x+1=0}}} , {{{a=1}}} , {{{b=-3}}} , and {{{c=1}}} , so
{{{x = (-(-3) +- sqrt((-3)^2-4*1*1 ))/(2*1) }}} ,
{{{x = (3 +- sqrt(9-4))/2}}} ,
{{{x = (3 +- sqrt(5))/2}}} .
That is the same as
{{{x = 3/2 +- sqrt(5)/2}}} , which is a way of saying {{{system(highlight(x=3/2+sqrt(5)/2),"or",highlight(x=3/2-sqrt(5)/2))}}}