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Re: abc < 0 and b^2c > 0
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01 Sep 2018, 15:02
We have \(abc<0\) this means that either a, b , c or all three are negative. Since product of the three is negative.
Now \(b^2c>0\). Since \(b^2\) is always a positive number then \(c\) must be a positive number for \(b^2c>0\) to be true.
For \(abc<0\) to be true \(ab<0\) must be true, since c is a positive number.
Thus Quantity A, namely \(ab\), is less than Quantity B, which is 0.