Question

If α and β are the roots of the equation x² + px + 2 = 0 and `1/α` and `1/β` are the roots of the equation 2x² + 2qx + 1 = 0, then `(α - 1/α)(β - 1/β)(α + 1/β)(β + 1/α)` is equal to ______.

Options

• `9/4(9 + q)^2`

• `9/4(9 - q)^2`

• `9/4(9 + p^2)`

• `9/4(9 - p^2)`

Answer 1

If α and β are the roots of the equation x² + px + 2 = 0 and `1/α` and `1/β` are the roots of the equation 2x² + 2qx + 1 = 0, then `(α - 1/α)(β - 1/β)(α + 1/β)(β + 1/α)` is equal to `underlinebb(9/4(9 - p^2))`.

Explanation:

α.β and α + β = –p also `1/α + 1/β` = –q

⇒ p = 2q

Now `(α - 1/α)(b - 1/β)(α + 1/β)(β + 1/α)`

= `[αβ + 1/(αβ) - α/β - β/α][αβ + 1/(αβ) + 2]`

= `9/2[5/2 - (α^2 + β^2)/2]`

= `9/4[5 - (p^2 - 4)]`

= `9/4(9 - p^2)`  ...[∵ α² + β² = (α + β)² –2αβ]