Irrational inequality. Symbolab and Wolfram Alpha have different answers

Question

Question :

Find the solution for x!

$\dfrac{\sqrt{x^2-4}}{x-2}\ge 0$

1. For my own work, i found :

$x<-2 \quad \text{or}\quad x>2$

2. On Wolfram Alpha app, i type this command

(sqrt(x^2 - 4))/(x-2)>=0

And give me the answer :

$x=-2 \quad \text{or}\quad x>2$

3. On symbolab app, it gives me the answer :

$x\le-2 \quad \text{or}\quad x\ge2$

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What is wrong? Please somebody help me, which one is true? Or what is the correct answer if they all are wrong. And how to do it? Cz i do with basic way for solving this inequality. Squaring, and define the domain.

Answer 1

The square root is always positive or $0$. So the denominator of the fraction $\dfrac{\sqrt{x^2-4}}{x-2}\ge 0$ is positive only when $x>2$ (when $x=2$ is impossible). For the numerator, you have to take the C.E. (or existance condition) so $x^2-4\geq 0$: this is true for $x\leq-2$ or $x\geq 2$. Combinig the two condition, I obtain: $x>2$.

Answer 2

$\sqrt{x^2-4}$ is not negative. If it is $0$, the inequality holds; otherwise, the denominator must be positive. By the other hand, $x\neq 2$ to avoid 'division by zero'.

The correct solution is Wolfram Alpha's.

Answer 3

$x =-2$ is one solution. For $x \neq -2$ the ratio is positive only when the denominator is positive. So $x=-2$ or $x >2$.

Answer 4

I agree with Wolfram Alpha. $\dfrac{\sqrt{x^2-4}}{x-2}$ is defined when $|x|\ge2$ (because of the numerator)

and $x\ne2$ (because of the denominator).

It is positive when $x>2$ (because then both numerator and denominator are positive),

zero when $x=-2$ (because then numerator is zero and denominator is non-zero),

and negative when $x<-2$ (because then numerator is positive but denominator is negative).