$x^2 +2ax+b=0$ 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 .

@KaviRamaMurthy Please tell which one is not clear . I’ll define — Rider, Apr 11 '21 at 12:11
@Rider You said that the maximum value of it is $2m$. That doesn’t mean it is equal to it. — Vishu, Apr 11 '21 at 12:27

Siong Thye Goh · Answer 1 · 2021-04-11T12:29:36.370

1

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

edited Apr 11 '21 at 12:29

answered Apr 11 '21 at 12:15

Siong Thye Goh

149,520
20
88
149

Ok. The answer is that b belong to (a^2-m^2 , a^2] – Rider Apr 11 '21 at 12:27

score 0 · Accepted Answer · answered Apr 11 '21 at 12:29

0

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

answered Apr 11 '21 at 12:29

Vishu

14,469

Discriminant is $ 4(a^2 - b)$ with sqrt.and not $a^2 - b$ Why does $a^2 - b $need to be +ve ? – Rider May 13 '21 at 07:55
Also , why have you written D > 0 and D less than equal to 2m. I got 1st one but why 2nd one ? – Rider May 13 '21 at 08:02
Discriminant doesn’t have sqrt. Why did you use it ? – Rider May 13 '21 at 08:06
A number is positive if and only if $4$ times it is positive. – Vishu May 13 '21 at 08:44
I told you in the comments. You said that the maximum value of the difference of roots is $2m$. – Vishu May 13 '21 at 08:45
So what? The difference of the roots involves a root. – Vishu May 13 '21 at 08:46

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

2 Answers2