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Solve the inequalities. Express your answer in interval notation. x^2-8x+12>0 Solve the inequalities. Express your answer in interval notation. x^2-8x+12>0 @Mathematics
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\[ x<2\text{ or }x>6 \]
x 2 −8x+12=0 D=64−48=16 x 1,2 =(8±4)/2=6;2 when x<2, or x>6 x 2 −8x+12>0 , so the interval is (2;6)
sorry again, \[(-\infty;2) and(6;\infty)\]
\[(-\infty,2)\cup(6,\infty)\]
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