<s>[INST] Proving that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$ without the quadratic formula and without calculus I'm asked to prove that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$  in an algebra textbook.
The formula for the real roots of a second degree polynomial is not introduced yet. And the book is written without assuming any prior calculus knowledge so I can't  prove this by finding the minimum and the limits as x approaches $\infty \text{ and} -\infty  $. 
So there has to be a simple algebraic proof involving neither the quadratic formula nor calculus but I'm stuck. 
Here are some things I thought:
Method 1:

$1+x+x^2 = 0 \iff 1+x+x^2+x = x$
$\iff x^2+2x+1 = x$
$\iff (x+1)^2 = x $ 
And here maybe prove that there is no x such that $(x+1)^2 = x$ ???

Method 2:

$1+x+x^2 = 0$
$\iff x^2+1 = -x$
By the trichotomy law only one of these propositions hold: $x=0$ or
  $x>0$ or $x<0$.
Assuming $x=0$:
$x^2+1= 0^2+1 = 0 +1 = 1$
$-x = - 0 = 0$
And $1\neq 0$
Assuming $x>0$:
$x>0 \implies -x < 0$
And $x^2+1 \ge 1 \text{  } \forall x$
With this method I have trouble proving the case $x<0$:
I thought maybe something like this could help but I'm not sure:
$x<0 \implies -x=|x|$ 
$x^2 = |x|^2$
And then prove that there is no x such that $|x|^2 + 1 = |x|$??

Can anyone please help me? Remember: No calculus or quadratic formula allowed.
 [/INST] To pursue your ideas:
"And here maybe prove that there is no $x$ such that $(x+1)^2=x$ ???"
If $x > 0$ then $x + 1 > x$ and $x + 1 > 1$ so $(x+1)^2 = (x+1)(x+1) > x *1 = x$.
If $x = 0$ then $(x+1)^2 = 1 \ne 0 = x$
If $x < 0$ then $(x+1)^2 \ge 0 > x$.
"With this method I have trouble proving the case x<0"
If $x > 0$ then $x^2 > 0; x> 0; 1 > 0$ so $1 + x + x^2 > 0$. 
If $x =0$ then $1 + x + x^2 = 1 > 0$.
So $x < 0$.
"And then prove that there is no x such that $|x|^2+1=|x|$"
$|x|^2 \ge 0$ so $|x|^2 + 1 \ge 1$ so if $|x|^2 + 1 = |x|$ then $|x| \ge 1$.
If $|x| = 1$ then $|x|^2 + 1 = 2 \ne 1 = |x|$.
If $|x| > 1$ then $|x|^2 > |x|$ so $|x|^2 + 1 > |x| + 1 > |x|$.
====
Basically the first thing to notice is that if $1 + x + x^2 = 0$ then $1>0; x^2 \ge 0$ so $1 + x^2 = -x \ge 1 > 0$ so $x$ is negative.
The next thing to notice is $1 + x^2 = |x|$ So $x^2 < |x|$ so $|x| < 1$.  But that contradicts $|x| = x^2 + 1 \ge 1$.
But another approach is:
$1 + x +x^2 = 0$
$x(x + 1) = -1$ so $x$ and $x+1$ must be opposite signs.
So either $x > 0$ and $x + 1 < 0$ which is impossible or $x < 0$ and $x +1 > 0$.
But if $x< 0 < x+1$ the $|x| < 1$ and and $|x+1| < 1$.  So $|x(x+1)| < 1$ which is a contradiction.
</s>