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. \(\alpha\gamma\)
• B. 0
• C. \(\beta\gamma\)
• D. \(\alpha\beta\)

Hint:

Since \(\beta^2 = \alpha\gamma\), the first equation has equal roots \(-\beta/\alpha\); this must satisfy \(x^2+x-1=0\).

Answer 1

The Correct Option is C

G.P. with ratio r: \(\beta = \alpha r\), \(\gamma = \alpha r^2\), so \(\beta^2 = \alpha\gamma\).
Discriminant of \(\alpha x^2+2\beta x+\gamma\) is \(4\beta^2 - 4\alpha\gamma = 0\), so it has a repeated root \(x = -\frac{\beta}{\alpha} = -r\).
This root is common with \(x^2+x-1=0\): \(r^2 - r - 1 = 0\), so \(r^2 = r+1\).
\(\alpha(\beta+\gamma) = \alpha^2(r + r^2) = \alpha^2(2r+1)\).
\(r^3 = r\cdot r^2 = r^2 + r = 2r+1\). So \(\alpha(\beta+\gamma) = \alpha^2 r^3 = \beta\gamma\) (since \(\beta\gamma = \alpha r\cdot\alpha r^2 = \alpha^2 r^3\)).
Answer: (3) \(\beta\gamma\)