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Question

Let a,b,cR and a0. If α is a root of a2x2+bx+c=0,β is a root of a2x2bxc=0 and 0<α<β, then the equation, a2x2+2bx+2c=0 has a root γ that always satisfies

A
γ=β
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B
α<γ<β
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C
γ=(α+β)/2
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D
γ=α
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Solution

The correct option is B α<γ<β
Let f(x)=a2x2+2bx+2c
From the question, a2α2+bα+c=0 and a2β2bβc=0
Now, f(α)=a2α2+2bα+2c=bα+c=a2α2f(β)=a2β2+2bβ+2c=3(bβ+c)=3a2β2
but 0<α<β
α and β are Positive.
f(α)<0, f(β)>0
Hence, α<γ<β.

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