Proving that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$ without the quadratic formula and without calculus

Question

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.

Answer 1

Clearly, if $x\ge 0$ then $x^2+x+1\ge 1>0$. And if $x<0$, then $x^2+x+1>x^2+2x+1=(x+1)^2\ge 0$

Answer 2

Notice that: $$ x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4} \geq 0+\dfrac{3}{4}= \dfrac{3}{4} \ \ .$$

Answer 3

$x^2+x+1=(x^2+x+\frac{1}{4})+\frac{3}{4}>(x+\frac{1}{2})^2$

Answer 4

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.

Answer 5

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.

Answer 6

$x=0$ 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.

Answer 7

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 . . .

Answer 8

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

Answer 9

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.

Answer 10

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.

Answer 11

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.

Answer 12

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:

1. First, note the parabola opens up because the coefficient of the $x^2$ is positive. This means the vertex will be the minimum point.
   
2. Find the $x$-coordinate of the vertex with the equation $x=-\frac{b}{2a}$. In our case, $x=-\frac{1}{2}$.
   
3. 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.

Answer 13

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

Answer 14

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 $|x|^2 + 1 > |x| + 1 > |x|$.

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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.