If ,, are three consecutive terms of a non-constant G.P.

Question

If $\alpha,\beta,\gamma$ 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:

Accepted Answer

Let the common ratio of G.P is r

Then the equation of  $$\alpha {x^2} + 2\beta x + \gamma  = 0$$

$$ \Rightarrow $$ $$\alpha {x^2} + 2\alpha rx + \alpha {r^2} = 0$$

$$ \Rightarrow $$ x2 + 2rx + r2 = 0 ........(i)

Equation (i) and x2 + x - 1 = 0 ......... (ii) has a common root.

Now (i) - (ii) $$ \Rightarrow $$ (2r-1)x + (r2+1) = 0

$$ \Rightarrow $$ x = $${{ - \left( {{r^2} + 1} \right)} \over {2r - 1}}$$ ....... (iii)

Now putting (iii) in equation (ii)

$$ \Rightarrow $$ (r2+1)2 - (r2+1) (2r - 1) -  (2r - 1)2 = 0

$$ \Rightarrow $$ r4 - 2r3 - r2 + 2r + 1 = 0 .......... (iv)

dividing equation (iv) by r2

$$ \Rightarrow $$ r2 - 2r - 1 + $${2 \over r} + {1 \over {{r^2}}} = 0$$

$$ \Rightarrow $$ $${\left( {r - {1 \over r}} \right)^2} - 2\left( {r - {1 \over r}} \right) + 1 = 0$$

$$ \Rightarrow $$ $${\left( {r - {1 \over r} - r} \right)^2} = 0$$

$$ \Rightarrow $$ $${{{r^2} - 1} \over r} = 1$$ $$ \Rightarrow $$ r2 = 1 + r

Now $$\alpha (\[\beta $$ + $$\gamma $$) $$ \Rightarrow $$ $$\alpha (\[\alpha $$r + $$\alpha $$r2)

$$ \Rightarrow $$ $$\alpha ^2 r (1 + r)$$   [$$ \because $$  r2 = 1 + r]

$$ \Rightarrow $$ $$\alpha ^2 r . r^2$$ $$ \Rightarrow $$ $$( \alpha r )\,( \alpha r^2 )$$ $$ \Rightarrow $$ $$\beta $$ $\gamma $

Solutions · 1