Originally Posted by
Soroban Hello, NameIsHidden!
Watch what happens when we expand . . .
$\displaystyle \quad\left(x+\frac{1}{x}\right)^2 - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;4$
$\displaystyle x^2 + 2 + \frac{1}{x^2} - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;4$
$\displaystyle x^2 - 2 + \frac{1}{x^2} - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;0$
$\displaystyle \left(x -\frac{1}{x}\right)^2 - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;0 $
Let $\displaystyle u \,=\,x+\frac{1}{x}\!:\;\;u^2 - \tfrac{3}{2}u \:=\:0 \quad\Rightarrow\quad u\left(u - \tfrac{3}{2}\right) \:=\:0$
Hence: .$\displaystyle u \,=\,0,\;u \,=\,\tfrac{3}{2}$
Back-substitute:
. . $\displaystyle x + \frac{1}{x} \:=\:0 \quad\Rightarrow\quad x^2-1\:=\:0 \quad\Rightarrow\quad \boxed{x \:=\:\pm1}$
. . $\displaystyle x + \frac{1}{x} \:=\:\frac{3}{2} \quad \Rightarrow\quad 2x^2 - 3x - 2 \:=\:0 \quad\Rightarrow\quad (x-2)(2x+1) \:=\:0 $
. . . . . . $\displaystyle \boxed{x \;=\;2.\:\text{-}\tfrac{1}{2}}$