Let f(x)=ax2+bx+c be a quadratic equation with roots α,βf(2)+f(−1)=0⇒5a+b+2c=0
Also f(3)=0⇒9a+3b+c=0
⇒−5a=13b=6c
Product of roots αβ=ac=−56
We know α=3⇒β=−52∈(−1,0)
Let f(x) be a quadratic polynomial such that f(–1)+f(2)=0. If one of the roots of f(x)=0 is 3, then its other root lies in
Held on 2 Sept 2020 · Verified 6 Jul 2026.
(−1,0)
(1,3)
(–3,–1)
(0,1)
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