The complete solution of the in equation $x^{2} - 4 x < 12$ is

x < -2 or x > 6

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-6 < x < 2

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2 < x < 6

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-2 < x < 6

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Solution

The correct option is D
-2 < x < 6

$x^{2} - 4 x < 12$

⇒ $x^{2} - 4 x - 12 < 0$ ⇒ $x^{2} - 6 x + 2 x - 12 < 0$

⇒ (x-6)(x+2) < 0 ⇒ -2 < x < 6

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Similar questions

The complete solution of the in equation $x^{2} - 4 x < 12$ is

x^{2} + 4 x + 6 > 0

t a n^{- 1} \frac{1}{2 x + 1} + t a n^{- 1} \frac{1}{4 x + 1} = t a n^{- 1} \frac{2}{x^{2}}

\frac{d y}{d x} - y (\frac{x^{4} + 3 x^{2}}{(x^{2} + 1)^{2}}) = (4 x + 3) e^{\frac{x^{3}}{x^{2} + 1}}

y = f (x)

f (0) = 1

f (1) + f (- 1)

b

2 {log}_{\frac{1}{25}} (b x + 28) = - {log}_{5} (12 - 4 x - x^{2})

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