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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

quadratic equation

Factorization

Check the Answer


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

quadratic equation

Factorization

Check the Answer


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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