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

Solution 1:

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!)

Solution 2:

Because you know $x$ must satisfy all three inequalities, your conclusion would prove that no triangle exists, so you should be able to answer the question.

Alas, it turns out your conclusion is not true; try solving the inequalities again, and seeking some way to verify whether or not your solutions are correct.