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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 ?

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  • $\begingroup$ If the discriminant is negative, there are no real roots, no matter what $a$ is so in your case, the domain of your radical term is all real numbers and the outcome is indeed always positive $\endgroup$
    – imranfat
    Commented Sep 19, 2014 at 15:36
  • $\begingroup$ $x^4-7x^2+16=\left(x^2-\frac 72\right)^2+\frac{15}{4}.$ $\endgroup$
    – mathlove
    Commented Sep 19, 2014 at 15:37
  • $\begingroup$ @imranfat Here what I want to prove is that my original function is always positive, so $a$ matters, doesn't it ?! $\endgroup$ Commented Sep 19, 2014 at 15:38
  • $\begingroup$ No, if the discriminant is negative, then there are no real roots!. Now that either means only positive outcomes OR negatives. In your case it is positive because $a>0$ $\endgroup$
    – imranfat
    Commented Sep 19, 2014 at 15:44
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    $\begingroup$ You can also write $$x^4-7x^2+16=x^4-8x^2+16+x^2=(x^2-4)^2+x^2$$ which is always strictly positive for real $x$. $\endgroup$ Commented Sep 19, 2014 at 15:45

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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!!

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    $\begingroup$ Ok! Thanks a lot! I'm reassured ! $\endgroup$ Commented Sep 19, 2014 at 15:43
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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 $y$.

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Since $$x^4-7x^2+16>x^4-8x^2+16=(x^2-4)^2$$ we have $$\sqrt{x^4-7x^2+16}>\sqrt{x^4-8x^2+16}=\sqrt{(x^2-4)^2}=|x^2-4|\ge 0.$$

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