If α, β and γ are three consecutive terms of a non-constant G.P. such that the equations αx^{2} + 2βx + γ = 0 and x^{2} + x – 1 = 0 have a common root, then α(β + γ ) is equal to :
(1) αγ (2) 0 (3) α, β (4) βγ
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 \(\alpha (\beta + \gamma) = \alpha^{2} (t + t^{2})\) \(Option \ 4\ \beta \gamma = \alpha t . \alpha t^{2} =\alpha^{2} t^{3} = a^{2} (t^{2} + t)\)