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Im solving inequality x^2+3x-4<0, and i need to put it on a graph. I know that the answer is -4
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pretty sure that is not the answer
ok, im wrong, -4<x<1
\[x^2+3x-4=(x+4)(x-1)\] zeros are at \(-4\) and \(1\) and since this is a parabola that opens up, it is negative between the zeros and positive outside them, so \[x^2+3x-4<0\] on the interval \((-4,1)\)
thank you! finally done with my assignment o.O
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