Reuben Brennan

2022-04-01

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

### Answer & Explanation

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$\frac{\left(a-b\right)\left(2a-b\right)}{a\left(a-b+c\right)}=\frac{\left(1+s\right)\left(2+s\right)}{1+s+p}$
where $s=-\frac{b}{a}$ is the root sum and $p=\frac{c}{a}$ is the root product.
$s\in \left[0,2\right]$ and for a fixed s, p attains a maximum of $\frac{{s}^{2}}{4}$ (when the roots are equal) and a minimum of 0 (if $s\le 1$) or $s-1$ (if $s\ge 1$). Accordingly, to minimise the expression we must $p=\frac{{s}^{2}}{4}$, whereupon it becomes
$4\cdot \frac{\left(1+s\right)\left(2+s\right)}{{\left(s+2\right)}^{2}}=4\cdot \frac{1+s}{2+s}$
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 $s\in \left[0,1\right]$ we set $p=0$ and the expression simplifies to $2+s$, whose maximum is 3.
- If $s\in \left[1,2\right]$ we set $p=s-1$ and the expression simplifies to $\frac{\left(1+s\right)\left(2+s\right)}{2s}=\frac{1}{s}+\frac{3}{2}+\frac{s}{2}$, which is decreasing on $\left[1,\sqrt{2}\right]$ 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].

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