# 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.

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

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.

• This is really nice! You might add why the odd case doesn't work (I can see but not sure a lot of readers at the level of sophistication of this question would). – Silverfish Aug 20 '15 at 9:02
• It turns out that for odd n the only real root is -1. I'll add the proof of that. – marty cohen Aug 20 '15 at 15:48

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)

• What does it mean to "scale" a function? – mysatellite Aug 20 '15 at 13:38

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.

• Why does it matter if it's positive? I thought I had to prove that the discriminant was less than zero. – Jonathan Aug 19 '15 at 19:30
• @Jonathan That's a question of understanding what the discriminant is telling you. If you prove the function is always positive then it is never zero. And except for the case where $r=0$ when you get $p(x)=ax^2$ where $a$ is the first term of the progression, the quadratic here is always positive. The case $r=0$ is anomalous, but not excluded by the question. A negative discriminant for a quadratic tells you that it is always positive or always negative depending on the sign of the coefficient of $x^2$. – Mark Bennet Aug 19 '15 at 19:35
• I understand now. Thank you. – Jonathan Aug 19 '15 at 19:37

The discriminant involves no $x$ !. And it is, I guess, $a^{2}r^{2}-4a^{2}r^{2}=-3a^{2}r^{2}$.

• Thanks, that was a dumb mistake! I'm surprised I didn't see that I used x in the discriminant. – Jonathan Aug 19 '15 at 19:26

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.