If $x^{2} + 6 x - 27 > 0$ and $x^{2} - 3 x - 4 < 0$ , then

x > 3

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x < 3

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3 < x < 4

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x = 7 / 2

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Solution

The correct option is C $3 < x < 4$
$\begin{matrix} x^{2} + 6 x - 27 > 0 x^{2} + 9 x - 3 x - 27 > 0 x (x + 9) - 3 (x + 9) > 0 (x - 3) (x + 9) > 0 x \in (- \infty, - 9) \cup (3, \infty) x^{2} - 3 x - 4 < 0 x^{2} - 4 x + x - 4 < 0 x (x - 4) + (x - 4) < 0 (x + 1) (x - 4) < 0 x \in (- 1, 4) ∴ x \in (3, 4) o r, 3 < x < 4 \end{matrix}$

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Similar questions

x^{2} + 6 x - 27 > 0

- x^{2} + 3 x + 4 > 0

x

\frac{2 x^{2} - 6 x + 4}{4 x^{2} - 10 x + 3}

\frac{x^{2} - 3 x}{2 x^{2} - 5 x}

x \neq

A = \int \frac{d x}{x^{2} + 6 x + 25}

B = \int \frac{d x}{x^{2} - 6 x - 27}

12 (A + B) = λ . {tan}^{- 1} (\frac{x + 3}{4}) + μ . ln ∣ ∣ ∣ \frac{x - 9}{x + 3} ∣ ∣ ∣ + C

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x^{2} {(x - 14)}^{2} =

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