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Solve the inequality \[\frac{4}{x^2+4}<0\] and express the solution as an interval.
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numerator is always positive, and so is the denominator, so no solution
Yeah I knew there is no way possible this can be less than 0....
\(4>0\) and \(x^2\geq 0\) for all \(x\) and so \(x^2+4\geq 4\) for all \(x\) and therefore \(\frac{4}{x^2+4}>0\) for all x
Thank you!!!
yw
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wait, how'd you get that last part?
what last part?
how is \[x^2 \ge 0\]
because it is a square
if you square a positive number, the answer is positive if you square a negative number, the answer is also positive the only way to get something not positive is to square 0 that is why \(x^2\geq0\)
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Oh... okay, wasn't reading it properly... thanks once again :)
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