can we have a triangle with sides $1, x$ and $x^2$? Can we have a triangle with sides $1$, $x$ and $x^2$? And what could $x$ be?
I try to approach this question by making 3 inequalities.
$1+x>x^2$,
$1+x^2>x$,
$x^2+x>1$  
and they come with different quadratic inequalities
$x^2-x-1<0$ (solution is $(1+\sqrt 5)/2 > x > (1-\sqrt 5)/2$ ) ;
$x^2-x+1>0$ (solution is $x > (-1+\sqrt 5)/2$ or $x < (-1-\sqrt 5)/2$ )
$x^2+x-1>0$ (no real solution)
Then I start to struggle with the next step.... Thank you

Your approach is fine.  $1+x > x^2$ has the solution set you have indicated, but since $x$ is the side of a triangle, we cannot have negative $x$.  Thus $0 < x < \frac{1 + \sqrt{5}}{2}$.  The second inequality, $1 + x^2 > x$ is satisfied by all real numbers, so no further restriction here.  The solutions set of $x^2 + x > 1$ is $x < \frac{-1-\sqrt{5}}{2}$ or $x > \frac{-1+\sqrt{5}}{2}$, and only positive $x$ matters.  Thus, any $x$ such that:
$$ \frac{-1+\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2} $$
will give a valid triangle with sides $1$, $x$ and $x^2$.  Approximately, $.618 < x < 1.618$.  (Coincidentally, the larger value is the golden ratio, and the lower value is its multiplicative inverse!)
