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.
 A: As a different sort of argument:  
Note that $$x^3-1=(x-1)(x^2+x+1)$$
Thus any root of $x^2+x+1$ is a cube root of $1$.  But the only real $n^{th}$ roots of $1$ are $\pm 1$ so the quadratic can have no real roots.  This last point is fairly clear, but just in case:   $|x|>1\implies \lim_{n\to \infty} |x|^n=\infty$ and $|x|<1\implies \lim_{n\to \infty} |x|^n=0$
As an alternative way to finish, note that $x^3-1$ has one real root, namely $x=1$, and it is montone increasing so it can not have another.
A: If 
$x^2 + x + 1 = 0, \tag 1$
then
$x^2 + x + \dfrac{1}{4} = -\dfrac{3}{4}; \tag 2$
but
$(x + \dfrac{1}{2})^2 = x^2 + x + \dfrac{1}{4}, \tag 3$
so
$(x + \dfrac{1}{2})^2 = -\dfrac{3}{4}; \tag 4$
but no real has a negaitive square, so . .  . 
A: Here is a variation on what others have written. It isn't really a different answer, but illustrates a useful trick. If $x^2+x+1=0$ then $$4x^2+4x+4=(2x+1)^2+3=0$$
We have $(2x+1)^2\ge 0$  and $3\gt 0$ so $(2x+1)^2+3\gt 0$. And depending on your axioms/definitions this is either an immediate contradiction or can be made into one.
A: To me, the simplest approach would be to graph the parabola and note it doesn't cross the $x$-axis. This is equivalent to stating it has no real roots.
If you need an algebraic proof, find the minimum:


*

*First, note the parabola opens up because the coefficient of the $x^2$ is positive. This means the vertex will be the minimum point.

*Find the $x$-coordinate of the vertex with the equation $x=-\frac{b}{2a}$. In our case, $x=-\frac{1}{2}$.

*Find the value of the function at the vertex. For us, this is $y=\frac{3}{4}$. 


So the minimum point of the function is $\left(-\frac{1}{2},\frac{3}{4}\right)$, which is above the $x$-axis, meaning the function has no real roots.
A: 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.
A: $x^2+x+1=(x^2+x+\frac{1}{4})+\frac{3}{4}>(x+\frac{1}{2})^2$
A: Method 1: Suppose $(x+1)^2 = x$. Certainly $x+1 > x$, so in order for this to be possible, we need $x+1 < 1$ (otherwise squaring it would make it even bigger), so $x < 0$. But then we have $(x+1)^2 = x$, a nonnegative number equal to a negative number.
A: Assume $\exists$ a $\in \mathbb{R}$ such that $a^2 + a + 1 = 0$. This implies $a^2 = -(a + 1)$. 
Look at these 3 exhaustive cases:
(1) $a + 1 \gt 0$ or $a > -1 $ is impossible since $a^2$ is non-negative.
(2) $a = -1$ is impossible since $1 \neq 0$
(3) $a < -1 \implies a^2 = |a| - 1$. Since $a^2 > |a|$ when $|a| > 1$ this is impossible
A: Another approach:
Suppose $r$ is a root. Clearly $r\neq 0$, and so
$$ 0 = 1+r+r^2 = r^2\left(1+\frac{1}{r}+\frac{1}{r^2}\right)$$
so the other root has to be $1/r$. By the fundamental theorem of algebra it follows that
$$ 1+x+x^2 = A(x-r)(x-1/r) = A\left(x^2 -(r+1/r)x + 1\right)$$
Now by equating coefficients we find that $A=1$ and $r+1/r = -1$, but for every $r<0$ and $r\neq -1$ either $r<-1$ or $1/r<-1$, so $r+1/r < -1$. Hence no such $r$ may be found.
A: In your Method 1, where you reach the equality (x + 1)² = x, it can be easily shown to never satisfy on R:
1) should it hold true, x is non-negative;
2) then x + 1 is not less than 1;
3) x + 1 > x ≥ 0, x + 1 ≥ 1, ⇒
(x + 1)(x + 1) > x · 1 = x
A: Notice that: 
$$
x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4} \geq 0+\dfrac{3}{4}= \dfrac{3}{4} 
\ \ .$$
A: Clearly, if $x\ge 0$ then $x^2+x+1\ge 1>0$. And if $x<0$, then $x^2+x+1>x^2+2x+1=(x+1)^2\ge 0$
A: $x=0$ is not a root, so divide by $x \ne 0$ and write the equation as:
$$
x+\frac{1}{x} = -1
$$
This requires $x$ to be negative for the LHS to be negative, but then $y=-x$ is positive and $\displaystyle y+\frac{1}{y} \ge 2$ by AM-GM, so $\displaystyle x+\frac{1}{x} \le -2 \lt -1\,$, therefore there are no real solutions.
A: Here's another way that doesn't involving completing the square (which is really just the quadratic formula done from first principles). If $1 + x + x^2 = 1 + x(1 + x) = 0$, then $x(1 + x)$ is negative, which implies that $-1 < x < 0$ and $0 < x + 1 < 1$, so $|x(1 + x)| < 1 $ and $x(1 + x) > -1$. Hence $1 + x + x^ 2 = 1 + x(1 + x) > 1 - 1 = 0$ giving a contradiction.
