Trying to figure out the method for solving [(x-9)(x+1)]/(x-2) greater than or equal to 0

Question

saifoo.khan · Answer

$$\frac{(x-9)(x+1)}{x-2} \ge 0 $$

saifoo.khan · Answer

$$(x-9)(x+1)\ge 0 $$

saifoo.khan · Answer

$$x \ge 9 , x \le - 1$$

anonymous · Answer

Thanks.  How do you express that in interval notation?

myininaya · Answer

saifoo you can't do that since x-2 is also sometimes negative

saifoo.khan · Answer

myininya, quick question, is -3 and -6 complex?

myininaya · Answer

$$f(x)=\frac{(x-9)(x+1)}{x-2} $$
f=0 when x=9, =-1
f is undefined at x=2

so we need to look around each of these numbers

------|--------|----------|---------
       -1              2                9
Test all 4 of these intervals.

myininaya · Answer

yes anything in a+bi form is complex
b can be 0

myininaya · Answer

where a and b are real

saifoo.khan · Answer

O I C.

saifoo.khan · Answer

So my answer is incorrect right?

myininaya · Answer

$$f(-2)=\frac{(-2-9)(-2+1)}{-2-2}=\frac{(-)(-)}{-}=-<0$$ $$f(0)=\frac{(-)(+)}{(-)}=(+)>0$$ $$f(3)=\frac{(-)(+)}{(+)}=(-)<0$$ $$f(10)=(+)>0$$ so f>=0 when...

myininaya · Answer

$$x \in [-1,2) \cup [9,\infty)$$