Let $S$ denotes the set of real values of $' a'$ for which the roots of the equation $x^{2} - a x - a^{2} = 0$ exceeds $S$ then $' a'$ equals to

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null set

$(\frac{- 1}{2}, \frac{1}{4})$

$(- \infty, 0)$

$(- 2, \frac{- 1}{2})$

answer is D.

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Detailed Solution

$\begin{array}{l} D \geq 0 \Rightarrow a^{2} + 4 a^{2} \geq 0 \\ f (a) > 0 \Rightarrow - a^{2} > 0 \\ \frac{- b}{2 a} > a \Rightarrow \frac{a}{2} > a \end{array}$

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