Question

★Suppose a,b are integers and a+b is a root x² + ax + b = 0. What is the maximum possible value of b²?★

Answer 1

$({a + b)}^{2} + (a + b)a + b = 0$
a2+b2+2ab+a2+ab+b=0
solved The equ b==?
then!b2 =

Answer 2

Answer:

Step-by-step explanation:

Given: a,b are integers. x² + ax +b=0

(a+b) is a root of it.

(a+b)²+a(a+b)+b=0

Or, b²+(3a+1)b+2a²=0

Or, b={ -(3a+1)+- ²V (3a+1)² - 8a}/2

SOLVING THIS WE GET,

SINCE a,b are integers, a²+ba+1 must be a perfect square and a(a+6) is also a perfect square.

Possible value of a=0, b=-6

If a=0, b=-1 or 0.

If a=-6, b=9 or 8

MAX POSSIBLE VALUE OF b² is 81

Hope this would help