Prove that the coefficients of a quadratic function with real roots cannot be in geometric progression

Note: I added a proof that for odd $n$ the only root is $-1$.

Generalizing Milo Brandt's answer, which I thought of before I saw his, this applies to a polynomial of any even degree.

If the polynomial is of degree $2n$, using his argument, we need to find out how many real roots $p(x) =x^{2n}+x^{2n-1}+...+x+1 $ can have.

But $p(x) =\frac{x^{2n+1}-1}{x-1} $ has no real roots because the numerator and denominator have the same sign and at 1, their common root, $p(x) = 2n+1$.

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For odd $n$, the only real root is $-1$. $n=3$ shows what happens; I will then give the proof for general odd $n$.

$x^3+x^2+x+1 =\frac{x^4-1}{x-1} =\frac{(x^2+1)(x^2-1)}{x-1} =\frac{(x^2+1)(x+1)(x-1)}{x-1} =(x^2+1)(x+1) $ for $x \ne 1$. The only real root is, obviously, $x=-1$.

For general odd $n$, since $n+1$ is even, let $n+1 = 2^km$ where $m$ is odd. Then, just for $n=3$, above,

$\begin{array}\\ x^n+x^{n-1}+...+x+1 &=\frac{x^{n+1}-1}{x-1}\\ &=\frac{x^{2^km}-1}{x-1}\\ &=\frac{(x^{2^{k-1}m}+1)(x^{2^{k-1}m}-1)}{x-1}\\ &=\frac{(x^{2^{k-1}m}+1)(x^{2^{k-2}m}+1)(x^{2^{k-2}m}-1)}{x-1}\\ &=\frac{(x^{2^{k-1}m}+1)(x^{2^{k-2}m}+1)...(x^{2m}+1)(x^m+1)(x^m-1)}{x-1}\\ &=(x^{2^{k-1}m}+1)(x^{2^{k-2}m}+1)...(x^{2m}+1)(x^m+1)\frac{x^m-1}{x-1}\\ \end{array} $

Since $m$ is odd, as proved above, $\frac{x^m-1}{x-1}$ has no real roots. All the terms $x^{2^jm}+1$ for $j \ge 1$ are at least $1$ since the exponent is even. Finally, since $m$ is odd, $x^m+1$ has as its only real root $x=-1$. Therefore, the whole polynomial has $-1$ as its only real root.

Here's one slick method which occurs to me after noting that $$f(x)=ax^2+arx+ar^2$$ is a homogenous polynomial in $x$ and $r$. This lets us "scale" our function so that all the $r$'s included have equal degree. In particular: $$f(xr)=ar^2x^2+ar^2x+ar^2$$ And then we see a constant factor of $ar^2$ which we divide out $$\frac{f(xr)}{ar^2}=x^2+x+1$$ which has no real roots, so neither does $f(x)$ (noting that neither $a$ nor $r$ may be zero)

The discriminant is just: $\Delta = (ar)^2 - 4(ar^2)(a) = -3(ar)^2 < 0$ because neither $a$ nor $r$ are allowed to be $0$.