Suppose $$ax^2+bx+c$$ is a quadratic polynomial (where $a$, $b$ and $c$ are not equal to zero) that has real roots. Prove that $a$, $b$, and $c$ cannot be consecutive terms in a geometric sequence.
I tried writing the geometric sequence as $$a,\ b=ar,\ c=ar^2$$ and then substituting it back into the quadratic as $$ax^2+arx+ar^2$$ and then factoring and trying to prove that the discriminant was less than zero. But I ended up with $$r^2(x-2)(x+2)$$ which is not always less than zero. Any help would be appreciated.
Best Answer
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, $p(x) = 2n+1$.
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.