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x^2-2x-5>orequal to 3
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{{x >= 4}, {x <= -2}}
start with \[x^2-2x-8\geq 0\] then find the zeros. since this is a parabola facing up with will be positive outside them
(x-4)(x+2)>=0
\[x \le -2 \] \[x \ge +4\]
\[(x-4)(x+2)\geq 0\]wow we all got this one!
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inf?
oops i forgot to be brackets
(-inf,-2] U [4,inf)
\[\Huge \infty\]
\[(-\infty,-2)\cup (4,\infty)\]
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you mean to include -2 and 4
\[\Huge \frac{\infty}{\color{blue}{\text{saifoooooo!}}}\]
\[(-\infty,-2] \cup [4,\infty)\]
yeah yeah that is the second time i made that mistake tonight
\[(-\infty,-2] \cup [4,\infty)\]
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i win :)
as per usual
lol
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