Let a, b, c be real numbers, a≠0. If α is a root of a2x2+bx+c=0. β is the root of a2x2−bx−c=0 and 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=α+β2
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C
γ=α
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D
α<γ<β
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Solution
The correct option is Cα<γ<β Take f(x)=a2x2+2bx+2c Now, f(α)=2(a2α2+bα+c)−a2α2……(1) Since α is a root of a2x2+bx+c So (1) becomes −a2α2 Then f(β)=−2(a2β2−bβ−c)+4a2β2……(2)which similarly becomes f(β)=2a2β2 Since f(α)f(β)=−4α2β2a4<0 Hence the root of f(x) ie γ lies between α and β i.e. α<γ<β.