How to solve a quadratic inequality that acts like a quadratic equality?

Question

This will be largely a trivial question.

But how do I solve an inequality like this: $3x^4 - 4x^2 + 1>0$ ?

Of course, I can treat it like a quadratic inequality by saying $t=x^2$
So I can solve it like: $ 3t^2 - 4t +1 >0$
The solutions are: $t_1=1; t_2=\frac{1}{3}$
Or in termes of $x$: $x_1=1; x_2=-1; x_3=\frac{\sqrt{3}}{3}; x_4=-\frac{\sqrt{3}}{3}$

So if this were an equality then I had solved it as I found all the roots.
But what does all this mean for inequality?
How do I know where is this polynomial larger or smaller than $0$?

Answer 1

The inequality is trivial as you only have even powers, so

$$\forall\,x\in\Bbb R\;,\;\;x^4\ge0\;,\;\;x^2\ge 0\implies 3x^4+4x^2+1\ge 1>0$$

Answer 2

HINT:

Since the product of the $4$ bracket expressions is positive, this implies that:

1. Either all the bracket expressions are positive

OR

2. Any two bracket expressions are positive and the other two are negative.

OR

3. All the bracket expressions are negative.