The value of k so that the equation $2 x^{2} + k x - 5 = 0$ & $x^{2} - 3 x - 4 = 0$ may have one root common.

3, \frac{- 27}{4}

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- 3, \frac{- 27}{4}

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3, \frac{27}{4}

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- 3, \frac{27}{4}

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Solution

The correct option is B $- 3, \frac{- 27}{4}$

Let $α$ be a common root.
$2 α^{2} + k α - 5 = 0$ and $α^{2} - 3 α - 4 = 0$
Solving the two equations
$\frac{α^{2}}{- 4 k - 15} = \frac{- α}{- 8 + 5} = \frac{1}{- 6 - k} \Rightarrow 4 k^{2} + 39 k + 81 = 0 \Rightarrow k = - 3 o r - \frac{27}{4}$

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