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

There is a natural separation between $x\gt 1$ and $x\lt 1$. For if $x\gt 1$, then $x^2\gt x\gt 1$, so $x^2$ is th longest isde.
On the other hand, if $0\lt x\lt 1$, then $x^2$ is the baby of the family.
Case $x \gt 1$: If the inequality $x^2 \lt x+1$ holds, we will be OK. The roots of $x^2-x-1=0$ are $(1\pm \sqrt{5}/2$. So in our range, the desired inequality holds for $1\lt x\lt (1+\sqrt{5})/2$.
Case $0\lt x\lt 1$: Now we want $x^2+x\gt 1$. The roots of $x^2+x-1=0$ are $(-1\pm\sqrt{5})/2$. so in our range, the desired inequality holds if $x\gt (-1+\sqrt{5})/2$.
Of course $x=1$ is fine too. Putting the results together, we see that we can form a triangle precisely if $(-1+\sqrt{5})/2 \lt x \lt (1+\sqrt{5})/2$. 
Remark: Consider the numbers $1,x, x^2$.  Suppose that we can make a triangle with these sides. The sides can be rewritten as $x^2(1/x^2)$, $x^2(1/x)$, and $x^2(1)$. By scaling, we can make a triangle with these sides iff we can make a triangle with sides $(1/x^2,1/x,1)$. Putting $y=1/x$, we see that there is a triangle with sides $1,x,x^2$ iff there is a triangle with sides $1,y,y^2$. Thus there is a natural correspondence between our triangles that have shortest side equal to $1$ with the triangles that have longest side equal to $1$. We could have used this correspondence to get an instant bound on $x$ for the case $x\lt 1$ from the bound for the case $x\gt 1$: just take the reciprocal. 
