Try this beautiful problem from Algebra based on Quadratic equation.
Suppose a,b are integers and a + b is a root of \(x^2 +ax+b=0\).What is the maximum possible
values of \( b^2 \)?
Algebra
quadratic equation
Factorization
Answer:$81$
PRMO-2018, Problem 9
Pre College Mathematics
(‘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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