If $6$ lies between the roots of the equation $x^{2} + 2 (a - 3) x + 9 = 0$ then

a \in (- \frac{3}{4}, \infty)

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a \in (- \infty, - \frac{3}{4})

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a \in (- \infty, 0) \cup (6, \infty)

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a \in (- \frac{3}{4}, 6)

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Solution

The correct option is B $a \in (- \infty, - \frac{3}{4})$

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