Expert answers to "x2+2ax+b=0 where x1 and x2 are real solutions.For..."

zatajuxoqj

Answered question

2022-04-01

x^{2} + 2 a x + 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} | = 2 m

Tristatex9tw

Beginner2022-04-02Added 18 answers

Step 1
We need

a^{2} - b > 0

, that is

a^{2} > b

a^{2} - b \leq m^{2}

b \geq a^{2} - m^{2}

In summary,

b \in [a^{2} - m^{2}, a^{2})

Jaslyn Allison

Beginner2022-04-03Added 13 answers

Step 1
The discriminant must be positive, i.e.

a^{2} - b > 0

The maximum difference of the roots is 2m, and the minimum difference is zero when

a^{2} - b \to 0

. So

0 < 2 \sqrt{a^{2} - b} \leq 2 m

0 < a^{2} - b \leq m^{2}

- m^{2} \leq b - a^{2} < 0

a^{2} - m^{2} \leq b < a^{2}

Hence,

b \in [a^{2} - m^{2}, a^{2})

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