How do you solve $-x^4-16>0$ using a sign chart?

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Answer 1 · 2017-09-11T12:08:59

There are no real solutions.

We have:

$-x^4-16 gt 0$

Giving:

$x^4+16 lt 0$

And $x^4 ge 0 AA x in RR => x^4+16 ge 16 AA x in RR$

Hence there are no real solutions.

How do you solve -x^4-16>0-x^4-16>0 using a sign chart?