If $α and β$ are the zeroes of the polynomial $x^{2} + 4 x - 77,$ then the quadratic polynomial whose zeroes are $\frac{1}{α}$ and $\frac{1}{β}$ can be equal to

4 x^{2} - 3 x + 85

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4 x^{2} - 12 x + 85

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77 x^{2} - 4 x - 1

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77 x^{2} - 12 x - 1

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Solution

The correct option is C $77 x^{2} - 4 x - 1$
Given that $α and β$ are the zeroes of $x^{2} + 4 x - 77 .$

$\Rightarrow α + β = - 4, αβ = - 77$

$∴ \frac{1}{α} + \frac{1}{β} = \frac{α + β}{αβ} = \frac{- 4}{- 77} = \frac{4}{77}$

Also, $\frac{1}{α} \times \frac{1}{β} = \frac{1}{αβ} = \frac{- 1}{77}$

∴Required polynomial is $k (x^{2} - (\frac{1}{α} + \frac{1}{β}) x + (\frac{1}{α} \times \frac{1}{β})],$ where k is real

$= k (x^{2} - \frac{4}{77} x - \frac{1}{77}]$

$= k (\frac{1}{77} (77 x^{2} - 4 x - 1)]$

$= 77 x^{2} - 4 x - 1$ (for $k = 1)$
Thus, the quadratic polynomial whose zeroes are $\frac{1}{α} and \frac{1}{β}, can be 77 x^{2} - 4 x - 1 .$

Hence, the correct answer is option (c).

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