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# Quadratic Equation Problem | PRMO-2018 | Problem 9

Try this beautiful problem from Algebra based on Quadratic equation from PRMO 8, 2018. You may use sequential hints to solve the problem.

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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