Question:
If $\alpha, \beta$ and $\gamma$ are three consecutive terms of a non-constant G.P. such that the equations $\alpha x^{2}+2 \beta x+\gamma=0$ and $x^{2}+x-1=0$ have a common root, then $\alpha(\beta+\gamma)$ is equal to :
Correct Option: 1
Solution:
$\alpha x^{2}+2 \beta x+\gamma=0$
Let $\beta=\alpha t, \gamma=\alpha t^{2}$
$\therefore \alpha x^{2}+2 \alpha t x+\alpha t^{2}=0$
$\Rightarrow x^{2}+2 t x+t^{2}=0$
$\Rightarrow(x+t)^{2}=0$
$\Rightarrow x=-t$
it must be root of equation $x^{2}+x-1=0$
$\therefore \mathrm{t}^{2}-\mathrm{t}-1=0$ ................(1)
Now
$\alpha(\beta+\gamma)=\alpha^{2}\left(t+t^{2}\right)$
Option $1 \beta \gamma=\alpha t \cdot \alpha t^{2}=\alpha^{2} t^{3}=a^{2}\left(t^{2}+t\right)$
(from equation 1)