Question 1066714
I interpret the wording of your question as
find {{{x}}} such that
{{{x^2+(1/x)^2 = x+1/x+7/4}}}
Let us say that {{{x+1/x=y}}} .
Then, {{{y^2=x^2+(1/x)^2+2x(1/x)}}} ,
{{{y^2=x^2+(1/x)^2+2}}} , and
{{{x^2+(1/x)^2=y^2-2}}} .
With that, we can re-write the equation as
{{{y^2-2=y+7/4}}} .
Multiplying both sides of the equal sign times {{{4}}} , we get
{{{4y^2-8=4y+7}}} ,
which simplifies to
{{{4y^2-4y-15=0}}}
Either applying the quadratic formula, or factoring,
or "completing the square", we can find the solutions to be
{{{y=5/2}}} and {{{y=-3/2}}} .
{{{x+1/x=-3/2}}} does not have real solutions,
but we can find real solutions for
{{{x+1/x=5/2}}} .
Multiplying both sides of the equal sign times {{{2x}}} ,
we get the equivalent equation
{{{2x^2+2x=5}}}<--->{{{2x^2+2x-5=0}}} ,
whose solutions are
{{{highlight(x=2)}}} and (of course) {{{highlight(x=1/2)}}} .