Question

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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

If a+ b is a root then it satisfies the equation, means when we put (a+b) in place of X the expression should be 0.

let's put it then , X = a+b

(a+b)^2 + a(a+b) + b =0

=> a^2 + b^2 +2ab + a^2 + ab + b =0

=> 2a^2 + b^2 +3ab + b = 0

here a^2 >=0

b^2 >=0

therefore , adding also >=0

2a^2 + b^2 >=0

therefore,

b^2 >= -2a^2

since a^2>=0,

-2a^2 <=0,

hence max b^2 = 0

Answer 2

$\sf\star\bold\red{{QUESTION:-}}$

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

$\sf\star\bold\green{{ANSWER:-}}$

Given that a+b is a root of the equation x²+ax+b=0.

Then,

(a+b)²+a(a+b)+b=0 i.e,

2a²+3ab+b²+b=0

This is a quadratic equation in both a and b

it's discriminant,

9b²-8(b²+b)=b²-8b

should be a perfect square. Take

b²-8b=p² so, that

(b-4)²-p²=16

(b-4+p)=16

As b and p are integers therefore three cases possible .

★Case 1:-b-4+p=16,b-4-p=1

$\implies$b=25/2 not possible.

★case 2 :-b-4+p=8,b-4-p=2

$\implies$b= 9

★case 3 :-b-4+p=4,b-4-p=4

$\implies$b=8

hence, maximum value of b is 9 and maximum value of b² is 81.