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# Quadratic Equation Problem | PRMO-2018 | Problem 9

Try this beautiful problem from Algebra based on Quadratic equation.

## Algebra based on Quadratic equation PRMO Problem 9

Suppose a,b are integers and a + b is a root of $x^2 +ax+b=0$.What is the maximum possible
values of $b^2$?

• $49$
• $81$
• $64$

### Key Concepts

Algebra

Factorization

Answer:$81$

PRMO-2018, Problem 9

Pre College Mathematics

## Try with Hints

(‘a+b”) is the root of the equation therefore (“a+b”) must satisfy the given equation

Can you now finish the problem ..........

Discriminant is a perfect square

can you finish the problem........

Given that $‘a+b”$ is the root of the equation therefore $“a+b”$ must satisfy the given equation

Therefore the given equation becomes ……

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

$\Rightarrow a^2 +2ab+b^2+a^2+ab+b=0$

$\Rightarrow 2a^2 +3ab+b^2+b=0$

Now since “a” is an integer,Discriminant is a perfect square

$\Rightarrow 9b^2 -8(b^2+b)=m^2$ (for some $m \in \mathbb Z)$

$\Rightarrow (b-4)^2 -16=m^2$

$\Rightarrow (b-4+m)(b-4-m)=16$

Therefore the possible cases are  $b-4+m=\pm 8$, $b-4-m=\pm 2$,$b-4+m=b-4-m=\pm 4$

i.e b-4=5,-5,4,-4

$\Rightarrow b =9,-1,8,0$

Therefore $(b^2)_{max} = 81$

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Try this beautiful problem from Algebra based on Quadratic equation.

## Algebra based on Quadratic equation PRMO Problem 9

Suppose a,b are integers and a + b is a root of $x^2 +ax+b=0$.What is the maximum possible
values of $b^2$?

• $49$
• $81$
• $64$

### Key Concepts

Algebra

Factorization

Answer:$81$

PRMO-2018, Problem 9

Pre College Mathematics

## Try with Hints

(‘a+b”) is the root of the equation therefore (“a+b”) must satisfy the given equation

Can you now finish the problem ..........

Discriminant is a perfect square

can you finish the problem........

Given that $‘a+b”$ is the root of the equation therefore $“a+b”$ must satisfy the given equation

Therefore the given equation becomes ……

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

$\Rightarrow a^2 +2ab+b^2+a^2+ab+b=0$

$\Rightarrow 2a^2 +3ab+b^2+b=0$

Now since “a” is an integer,Discriminant is a perfect square

$\Rightarrow 9b^2 -8(b^2+b)=m^2$ (for some $m \in \mathbb Z)$

$\Rightarrow (b-4)^2 -16=m^2$

$\Rightarrow (b-4+m)(b-4-m)=16$

Therefore the possible cases are  $b-4+m=\pm 8$, $b-4-m=\pm 2$,$b-4+m=b-4-m=\pm 4$

i.e b-4=5,-5,4,-4

$\Rightarrow b =9,-1,8,0$

Therefore $(b^2)_{max} = 81$

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