# 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 $|x|^2 + 1 = |x|$??

• Hint: $x^3-1=(x-1)(x^2+x+1)$.
– lulu
Aug 29, 2017 at 19:35
• Actually, $\{x\in\Bbb{R} | 1+x+x^2 = 0\}$ equals $\emptyset$, but not $\{\emptyset\}$. Aug 29, 2017 at 19:37
• Lord Shark the Unkwon :Could you please elaborate a little more on this? Or is it something that requires a lot of set theory to be understood and therefore wouldn't fit in the comment section?
– IMK
Aug 29, 2017 at 19:56
• @IMK Essentially, you have solved the problem in the post. Just merge the methods 1 and 2: a contradiction for the case $x<0$ in Method 2 can be obtained by the calculation in Method 1, according to which $x\geq 0$. Aug 29, 2017 at 23:10
• Lord Shark the Unknown was pointing out that $\emptyset=\{\}$, the set with nothing in it. This is different than $\{\emptyset\}=\{\{\}\}$, the set with one thing in it, and that one thing happens to be $\emptyset=\{\}$. (Something that requires a lot of set theory is that $\{\}$ is a common definition of the number zero, while $\{\emptyset\}$ is a common definition of the number one.) Sep 3, 2017 at 19:16

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

• I prefer this method for there is no splitting into cases. Arguably one might say that this is using the quadratic formula given that the quadratic formula is derived by completing the square :-) Sep 1, 2017 at 7:29

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

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 . . .

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.

• I like this one more than the splitting into cases argument or the midway-proof-of-quadratic-theorem arguments Sep 3, 2017 at 18:02

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.

"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.

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.

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

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.

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

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

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.

$x=0$ 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.

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.

• Explanation for the downvote, please! Oh, of course, that won't happen, because you haven't got the bottle to own up to your opinions! Aug 31, 2017 at 23:45