Answers

2014-10-29T18:40:58+08:00
 x^2-4x-12>0 \\x^2-4x-12=0 \\ \\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{4-\sqrt{ (-4)^2-4\cdot 1\cdot  (-12) }}{2 }=\frac{4-\sqrt{16+48 }}{2}=\\\\=\frac{4-\sqrt{64 }}{2}= \frac{4-8}{2}=\frac{-4}{2}=-2\\\\ x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{4+\sqrt{ (-4)^2-4\cdot 1\cdot  (-12) }}{2 }= \frac{4+8}{2}=\frac{12}{2}=6\\\\ Answer : \ \ x < -2\ \ \  or \ \ x > 6

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2014-10-30T11:37:58+08:00
The number that will make the inequality true is
X must be greater than or equal to 7 ..
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