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If α, β and γ are three consecutive terms of a non-constant G.P. such that the equations  αx2 + 2βx + γ = 0 and x2 + x – 1 = 0 have a common root, then α(β + γ ) is equal to :

(1)    αγ  (2) 0 (3)  α, β  (4) βγ

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Correct option is 4) βγ

\(\alpha x^{2} + 2\beta x + \gamma = 0\)
Let \(\beta = \alpha t, \gamma = \alpha t^{2}\)
\(\therefore \alpha x^{2} + 2\alpha tx + \alpha t^{2} = 0\)
\(\Rightarrow x^{2} + 2tx + t^{2} = 0\)
\(\Rightarrow (x + t)^{2} = 0\)
\(\Rightarrow x = -t\)
It must be root of equation \(x^{2} + x -1 = 0\)
\(\therefore t^{2} - t - 1 = 0 .... (1)\)
Now
\(\alpha (\beta + \gamma) = \alpha^{2} (t + t^{2})\)
\(Option \ 4\ \beta \gamma = \alpha t . \alpha t^{2} =\alpha^{2} t^{3} = a^{2} (t^{2} + t)\)

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