Do I justify it well? I have a function that is of the form and I want to prove that it is always positive :
$$\sqrt{x^{4}-7x^{2}+16}$$
I say that I can study $x^{4}-7x^{2}+16$ by putting $X = x^{2}$, which gives me $X^{2}-7X+16$ so I find the discriminant of this : $\Delta=b^{2}-4ac = (-7)^{2}-4\times1\times16 = 49-64=-15$.
Because $\Delta < 0$, we can conclude that there are no roots to both $x^{2}-7x+16$ and $x^{4}-7x^{2}+16$.  
Can I assume that $ax^{4}+bx^{2}+c$ is always positive if $\Delta < 0$ and $a>0$ ? Or do I need to add more justifications ?
Right now, because of that, I will assume that my function is always positive.
To conclude, I say that the square root of a positive number is always positive, and so that $\sqrt{x^{4}-7x^{2}+16}$ is always positive.
Are my explanations correct ?

If the discriminant is negative, the curve $y=ax^2+bx+c$ never intersects the x-axis. So it is either positive or negative $\forall x$. You can check the value of $y$ at any $x$ to be sure of the sign of $y$.
