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 :

• A. $\beta \gamma$
• B. $0$
• C. $\alpha \gamma$
• D. $\alpha \beta$

Answer 1

Answer: (A)

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