Correct option is (3) \(\beta \gamma\)
From question, \(\alpha,\,\beta\) and \(\gamma\) are three consecutive terms of a non-constant G.P.
From question, the equations \(\alpha x ^ 2 + 2\beta x + \gamma = 0\) and \(x^2 + x - 1 = 0\) have common roots.
The condition for common roots is:
\(\frac {\alpha}{1} = \frac {2 \beta}{1} =\frac {\gamma}{-1} =\lambda\)
Which is:
⇒ \(\alpha = \lambda\)
⇒ \(\beta = \frac {\lambda}{2}\)
⇒ \(\gamma = -\lambda\)
Now, from question,
⇒ \(\alpha (\beta + \gamma)\)
On substituting the available values,
⇒ \(\alpha (\beta +\gamma) =\lambda (\frac {\lambda }{2}-\lambda)\)
⇒ \(\alpha (\beta + \gamma) = \frac {\lambda^2-2 \lambda ^2}{2}\)
⇒ \(\alpha (\beta + \gamma) = \frac {- \lambda ^2}{2}\)
Now, we need to consider the options:
Option (a) : 0 = 0
Option (b) : \(\alpha \gamma\)
\(\Rightarrow \alpha \gamma = \lambda(-\lambda) = -\lambda ^2\)
Option (c) : \(\beta \gamma\)
\(\Rightarrow \beta \gamma = \frac {\lambda}{2} (-\lambda) =- \frac {\lambda ^2}{2}\)
Option (d) : \(\alpha \beta\)
\(\Rightarrow \alpha \beta = \lambda (\frac {\lambda}{2}) =\frac {\lambda ^2}{2}\)
