Question

❤Hello Mates❤
Please Answer This Question.

Suppose a, b are integers and a + b is a root of x² + ax + b = 0. What is the maximum possible

value of b²?​

Answer 1

Answer:

81

Step-by-step explanation:

(a+b)²+a(a+b)+b = 0

a²+2ab+b²+a²+ab+b = 0

2a²+3ab+b+b² = 0

b²+b(3a+1)+2a² = 0

$b = \frac{-(3a+1) \pm \sqrt{(3a+1)^2 - 8a^2} }{2}$      [From Sridhar Acharaya's equation]

$b = \frac{-(3a+1) \pm \sqrt{9a^2 + 6a + 1 -8a^2} }{2}$

$b = \frac{-(3a+1) \pm \sqrt{ a^2 + 6a + 1} }{y}$

$b = \frac{-(3a+1) \pm \sqrt{ a(a+6) + 1} }{y}$

As a,b ∈1, a(a + 6) + 1 must be a perfect square

Possible values of a are a = 0 or a = –6

If a = 0, b = –1 or 0

If a = –6, b = 9 or 8

Maximum possible value of b² = 81