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Question

# If a,b,c are real numbers and a≠0. If α is a root of a2x2+bx+c=0 and β is a root of a2x2−bx−c=0 such that 0<α<β, then the equation a2x2+2bx+2c=0 has a root γ that always satisfies

A
γ=α+β2
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B
γ=αβ2
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C
γ=αβ2
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D
α<γ<β
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Solution

## The correct option is D α<γ<βα is a root of a2x2+bx+c=0 ⇒a2α2+bα+c=0⇒bα+c=−a2α2...............(1) Similarly, ⇒bβ+c=a2β2...............(2) Let f(x)=a2x2+2bx+2c Then, f(α)=a2α2+2bα+2c=−a2α2<0 ......from(1)f(β)=a2β2+2bβ+2c=3a2β2>0 ......from(2) Thus, f(x) is a polynomial function such that f(α)<0 & f(β)>0 Therefore, there exists γ satifying α<γ<β such that f(γ)=0

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