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

Question

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 ?

Answer 1

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.

Answer 2

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.