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Q.

The values of 'a' for which the roots of the equation $x^{2} + x + a = 0$ are real and exceed 'a' are

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a

$0 < a < \frac{1}{4}$

b

$a < \frac{1}{4}$

c

$a < - 2$

d

$- 2 < a < 0$

answer is C.

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Detailed Solution

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$x^{2} + x + a = 0$
Both roots $> a \Rightarrow Δ \geq 0, f (a) > 0, α + β > 2 a$
$\begin{array}{l} \Rightarrow 1 - 4 a \geq 0, a^{2} + 2 a > 0, - 1 > 2 a \\ \Rightarrow a \leq \frac{1}{4}, a < - 2 or a > 0, a < - \frac{1}{2} \\ \Rightarrow a < - 2 \end{array}$

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