Question :
Find the solution for x!
$\dfrac{\sqrt{x^2-4}}{x-2}\ge 0$
- For my own work, i found :
$x<-2 \quad \text{or}\quad x>2$
- On Wolfram Alpha app, i type this command
(sqrt(x^2 - 4))/(x-2)>=0
And give me the answer :
$x=-2 \quad \text{or}\quad x>2$
- On symbolab app, it gives me the answer :
$x\le-2 \quad \text{or}\quad x\ge2$
...........................................................
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.
$\endgroup$4 Answers
$\begingroup$$\sqrt{x^2-4}$ is not negative. If it is $0$, the inequality holds; otherwise, the denominator must be positive. By the other hand, $x\neq 2$ to avoid 'division by zero'.
The correct solution is Wolfram Alpha's.
$\endgroup$ 3 $\begingroup$$x =-2$ is one solution. For $x \neq -2$ the ratio is positive only when the denominator is positive. So $x=-2$ or $x >2$.
$\endgroup$ 3 $\begingroup$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: $x>2$.
$\endgroup$ 1 $\begingroup$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).
$\endgroup$ 3
