$x^2 +2ax+b=0$ where $x_1$ and $x_2$ are real solutions. For which value of $b$ function $f(b)=|x_1-x_2|$ reaches maximum if $|x_1-x_2|=2m$

Question

$x^2 +2ax+b=0$ ; | A-C| = 2m. Roots are A,C-real,distinct. Then , b belongs to ?

How I solved it till now :

Using formula of A-C I,e ALPHA - beta = $\sqrt{D}$/a

$\sqrt{4(a^2-b)}$ = 2m

So , after solving it. I got |$a^2-b|=m^2. $

I’m not able to solve after this.

Answer of b has to be in a or m terms .

Answer 1

The discriminant must be positive, i.e. $$a^2-b\gt 0$$ The maximum difference of the roots is $2m$, and the minimum difference is zero when $a^2 -b \to 0$. So, $$0\lt 2\sqrt{a^2-b} \le 2m \\ 0\lt a^2 -b \le m^2 \\ -m^2 \le b-a^2 \lt 0 \\ a^2-m^2 \le b\lt a^2$$

Hence, $$b\in[a^2-m^2,a^2)$$

Answer 2

We need $a^2-b > 0$, that is $a^2 > b$.

$$a^2-b\le m^2$$

$$b\ge a^2-m^2$$

In summary, $$b \in [a^2-m^2, a^2)$$