The least positive integer value of $a$ for which both roots of the quadratic equation $(a^{2} - 6 a + 5) x^{2} + (\sqrt{a^{2} + 2 a}) x + (6 a - a^{2} - 8) = 0$ lie on either side of origin, is -

2

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1

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6

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3

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Solution

The correct option is D $3$
$a^{2} + 2 a \geq 0 \Rightarrow a \in (- \infty, - 2] \cup [0, \infty)$
Since, both the roots lie on either side of the origin.
Product of roots < 0
$\frac{c}{a} < 0$
$\Rightarrow \frac{6 a - a^{2} - 8}{a^{2} - 6 a + 5} < 0$
$\Rightarrow \frac{a^{2} - 6 a + 8}{a^{2} - 6 a + 5} > 0$
$\Rightarrow \frac{(a - 2) (a - 4)}{(a - 1) (a - 5)} > 0$
So $a \in (- \infty, - 2] \cup [0, 1) \cup (2, 4) \cup (5, \infty)$
So the least positive integer value of
$a = 3$

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