If value of $^{'} a^{'}$ for which the quadratic equation $3 x^{2} + 2 (a^{2} + 1) x + (a^{2} + 3 a + 2) = 0$ possesses roots of opposites sign lies in

(- \infty, 1)

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(- 2, - 1)

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(1, 2)

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(\frac{3}{2}, 2)

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Solution

The correct option is B $(- 2, - 1)$
We have,
$3 x^{2} + 2 (a^{2} + 1) x + (a^{2} + 3 a + 2) = 0 x^{2} + \frac{2}{3} (a^{2} + 1) x + \frac{1}{3} (a^{2} + 3 a + 2) = 0$

$I f r o o t s a r e o f o p p o s i t e s i g n p r o d u c t < 0$

$H e n c e, \frac{1}{3} (a^{2} + 3 a + 2) < 0 a^{2} + 3 a + 2 < 0 a (a + 2) + 1 (a + 2) < 0 (a + 1) (a + 2) < 0$

$H e n c e, f o r m t h e h e l p o f n u m b e r l i n e w e g e t a \in (- 2, - 1)$

Hence, this is the answer.

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