Range of an expression related to coefficients of a quadratic equation

Question

Let $ax^2+bx+c=0$ be a quadratic equation such that both the roots lie in [0,1]. What is the range of the expression $\frac{(a-b)(2a-b)}{a(a-b+c)}$ ?

I tried using the fact that $\frac{c}{a} \leq 1$ and tried to substitute $2a - b \geq a + b - c$, but got stuck

Answer 1

$$\frac{(a-b)(2a-b)}{a(a-b+c)}=\frac{(1+s)(2+s)}{1+s+p}$$ where $s=-\frac ba$ is the root sum and $p=\frac ca$ is the root product.

$s\in[0,2]$ and for a fixed $s$, $p$ attains a maximum of $s^2/4$ (when the roots are equal) and a minimum of $0$ (if $s\le1$) or $s-1$ (if $s\ge1$). Accordingly, to minimise the expression we must $p=s^2/4$, whereupon it becomes $$4\cdot\frac{(1+s)(2+s)}{(s+2)^2}=4\cdot\frac{1+s}{2+s}$$ and clearly this is minimised at $s=0$, i.e. the lower bound on range is $2$. To maximise the expression we consider two cases:

• If $s\in[0,1]$ we set $p=0$ and the expression simplifies to $2+s$, whose maximum is $3$.
• If $s\in[1,2]$ we set $p=s-1$ and the expression simplifies to $\frac{(1+s)(2+s)}{2s}=1/s+3/2+s/2$, which is decreasing on $[1,\sqrt2]$ and increasing otherwise. Thus we check the value at $s=2$, which is also $3$, and conclude that the maximum is still $3$.

Hence the expression range is $[2,3]$.