Conditions for roots of quadratic equation

Question

If I have a quadratic function $f(x)=ax^2+bx+c$, what are the conditions that should the numbers a,b and c satisfy so that the equation $f(x)=0$ has real roots $x_1$ and $x_2$ such that $x_1<6<x_2<10$?

My answer is the following system:

$f(6).f(10)<0$ (so that one of the roots is between 6 and 10)

$-\frac{b}{2a}<10.$

However, there is no such answer in the answer sheet.

Answer 1

The general criterion is that $\alpha$ separates the roots (their existence is presupposed) if an only if $af(x)<0$. Everything is based on this remark. If $af(\alpha)>0$ , $\alpha$ is either greater or smaller than both roots.

Here the conditions would be $a f(6)<0$ and $a f(10)>0$ plus checking whether $10$ is greater or smaller than both roots. But since $6$ is greater than one root, and $10>6$, $10$ can only be greater, so that there's nothing to check.

Finally, the conditions would be: $a f(6)<0$, $f(6)f(10)<0$.

The last condition alone only means that one of 6=$6, 10$ separates the roots, the other no, but you don't know which one. This ambiguity can also be raised with considerin the arithmetic mean of the roots, as you propose.

Answer 2

The requirements are as you did one part, and this gives the other part to complete it:

1. $f(6)\cdot f(10) < 0$, and

2. $\left|6+\dfrac{b}{2a}\right|< \dfrac{|x_2-x_1|}{2} \iff 36+\dfrac{6b}{a} + \dfrac{b^2}{4a} < \dfrac{(x_1+x_2)^2 - 4x_1x_2}{4}= \dfrac{b^2}{4a^2} - \dfrac{c}{a}$