Prove that the coefficients of a quadratic function with real roots cannot be in geometric progression 
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.
 A: 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)
A: The discriminant is just: $\Delta = (ar)^2 - 4(ar^2)(a) = -3(ar)^2 < 0$ because neither $a$ nor $r$ are allowed to be $0$.
A: Start with $ax^2+arx+ar^2$ and note first that the factor $a$ is irrelevant, so we may as well have $$p(x)=x^2+rx+r^2$$
Now this is a "known form" $x^2+xy+y^2=(x+\frac y2)^2+\frac 34y^2$ which is obviously positive unless $x=y=0$.
You do this by simply completing the square.
A: The discriminant involves no $x$ !. And it is, I guess, $a^{2}r^{2}-4a^{2}r^{2}=-3a^{2}r^{2}$.
A: Notice, we have $$ax^2+bx+c$$ Now, assume that $a$, $b$ & $c$ are in G.P. for which $ax^2+bx+c=0$ has real roots then we can take the values as $a=\frac{p}{r}$,  $b=p$ & $c=pr$ 
Where, $p\neq 0$ & $p, r>0$  
Now, setting the values of $a, b, c$, we have the following quadratic equation $$\frac{p}{r}x^2+px+pr=0$$ $$x^2+rx+r^2=0$$ Now, checking discriminant $$\Delta=B^2-4AC=(r^2)-4(1)(r^2)=-3r^2<0$$ This represents that the roots are imaginary. Hence, this is contradiction. 
Hence,  $ax^2+bx+c$  will have roots only then $a, b, c$ can't be in a G.P. 
A: 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.
