The difference of the roots of the quadratic equation $x^2 + bx + c = 0$ is $|b - 2c|$. If $c \neq 0$, then find $c$ in terms of $b$.
I know Vieta, for sum and product of roots of a quadratic, but am not seeing how to apply any of those tools to find the absolute difference $|b-2c|.$ I'm definitely missing a nice trick to do this. Solutions?
On
One could also approach the question in this way. We are given that the difference between the two zeroes of this (monic) quadratic polynomial is $ \ |b - 2c| \ \ . $ Since we cannot distinguish the relative "order" of the zeroes in the abstract, we may simply designate them $ \ r \ $ and $ \ r + b - 2c \ \ . $
The factorization of the polynomial is then $ \ (x - r)·(x - r - b + 2c) \ \ , $ thereby giving us $$ x^2 + bx + c \ \ = \ \ x^2 \ + \ (-b + 2c - 2r)·x \ + \ (r^2 + rb - 2rc) \ \ . $$
Consequently, $$ b \ = \ -b + 2c - 2r \ \ \Rightarrow \ \ 2b \ = \ 2c - 2r \ \ \Rightarrow \ \ b \ = \ c - r \ \ ; $$ $$ c \ = \ r^2 + rb - 2rc \ = \ r^2 + r·(c - r) - 2rc \ = \ r^2 + rc - r^2 - 2rc \ = \ -rc \ \ . $$
We find that $ \ c + rc \ = \ c·(1 + r) \ = \ 0 \ \ . $ Since the stated conditions exclude $ \ c = 0 \ \ , $ we conclude that $ \ r = -1 \ \ , $ and thus $ \ b \ = \ c - (-1) \ \Rightarrow \ c \ = \ b - 1 \ \ . $
Note that although the problem statement places no restrictions on the zeroes being real versus complex, we see that one and therefore both zeroes are real.
EDIT (9/26) -- After a little further investigation, I find that the relation also works if we permit $ \ b \ $ and $ \ c \ $ to be complex. The polynomial $ \ z^2 + (\alpha \ + \ \beta·i)·z + \ ([\alpha - 1] \ + \ \beta·i) \ $ has the one real zero $ \ -1 \ $ and one complex zero which differs by $ \ |b - 2c| \ \ . $
From the quadratic formula, one root is $\dfrac{-b+\sqrt{b^2-4c}}2$, and the other is $\dfrac{-b-\sqrt{b^2-4c}}2$,
so the difference is $\sqrt{b^2-4c}$.
So we have $b^2-4c=(b-2c)^2=b^2-4bc+4c^2$, or $4c^2+4c-4bc=0$.
If $c\ne0$, that means $4c+4-4b=0$. Can you take it from here?