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 there are no real roots and the sign of the polynomial is the same sign as $a$, on the whole interval.
So, since $X^2-7X+16$ is always positive and not equal to $0$, so also $x^4-7x^2+16$, and since it is known that the square root is greater or equal to zero, we have that $\sqrt{x^4-7x^2+16}>0$
So, it is correct as you said!!
