# How do I solve for $x$ in ${ ( x^2 + 2 ) ( x^2 - 16 )^{1/2} } \over { ( |x| + 2 ) ( x^2 - 9 ) }$ $\leq 0$?

My solving :

When $x\leq 0$

For $x^2 -16 \geq 0 \implies x \leq -4$

Checking for $x \leq -4$ , our expression , it will be true ( when the expression under square root gives positive output )

When $x\geq 0$

$x \geq 4$ so that the expression under square root is valid .

Checking expression , we get its true when ( expression under square root is taken negative )

But by this approach I get x belonging to $( -\infty, -4 ] \cup [ 4 , \infty )$ , but the answer is $\{ 4 , -4 \}$

My doubt 1 : Should I consider $( x^2 - 16 )^{1/2}$ To always give a positive output ? I seem to get { -4 , 4 } when I consider $\sqrt{x^2 - 16}$ $\geq 0$ . Should I always assume the $\sqrt{}$ to give positive output while solving equations ? are there special cases ?

• What do you mean by solve''? This isn't an equation. Are you finding the zeros of the function? – Axesilo Dec 13 '15 at 4:20
• @DA29731 I corrected it now . i forgot to mention " the expression $\leq 0$ " . – Ricky Dec 13 '15 at 4:21
• Oh, I see, it's an inequality. – Axesilo Dec 13 '15 at 4:22
• So , I figured out that the answer comes correct when I take $( x^2 - 16 )^{1/2}$ $\geq 0$ for all x real . – Ricky Dec 13 '15 at 4:33
• That's correct. Assuming you're doing pre-calculus, calculus, or real variables, you always have $\sqrt{x^2} = |x|$. (If you're working on complex analysis there's another story.) – Axesilo Dec 13 '15 at 4:40

It's actually very short:

1. $x^2+2$ and $|x|+2$ are strictly positive always so ignore them (i.e. multiply both sides by $(|x|+2)/(x^2+2)$)
2. $x^2$ needs to be greater than or equal to $16$ so that $(x^2-16)^{1/2}$ is well-defined (in your context) so $x^2-9>0$ so ignore that too
3. If we actually have $x^2>16$ then your fraction is strictly positive.

Together 2 and 3 imply that $x^2$ must be $16$. You then verify that $x=\pm 4$ is satisfactory.

Consider the two cases you have discerned: $x \geq 4$ and $x \leq -4$. I'll just do the first case.

You have a product of four terms: $(x^2 + 2)$, $\sqrt{x^2 - 16}$, $(|x| + 2)^{-1}$, and $(x^2 - 9)^{-1}$.

For each of these terms, whenever $x > 4$, the term is a positive number. (e.g. $x^2 - 9 > 4^2 - 9 = 7$).

So unless $x = 4$ exactly, you get a product of four positive numbers, so it can't be $\leq 0$.

Edit: A more general (calculus-style) approach to solving a problem like this is to notice that your function, call it $f(x)$, is continuous everywhere it is defined. So the only places $f(x)$ could switch from $\leq 0$ to $\geq 0$ or vice versa are places where $f(x) = 0$ or $f(x)$ is undefined.

Then you find the critical points where $f(x) = 0$ or is undefined. As you noticed, the function is undefined on the interval $(-4, 4)$ and defined everywhere else. Also, $f(x) = 0$ precisely when $x = -4$ or $x = 4$.

Then you can reason like this: $f(5)$ is positive, which means $f(x)$ is positive everywhere on the interval $(4, \infty)$, since it can't switch over to negative without hitting $0$ or being undefined somewhere.