Question

The set solution of the I equation:
| (X^2-3)/X | <= 2
Is:

Answer 1

When I do this exercise I found the set equals to [1,3], but, I think that is not the correct answer.

Answer 2

(X^2-3)/X <= 2 or (X^2-3)/X >= -2

Answer 3

Right,it's true ^^

Answer 4

But,what value you obtained?

Answer 5

actually it is an "and" statement

Answer 6

I know that I need to separate in two cases, but the result is not [1,3] ( I found this result)

Answer 7

What?

Answer 8

\[\frac{x^2-3}{x}\leq 2\]
\[\frac{x^2-3}{x}-2\leq 0\]
\[\frac{x^2-2x-3}{x}\leq 0\]
\[\frac{(x-3)(x+2)}{x}\leq 0\]
\(x\leq-3\) or \(0<x\leq3 \) and now you have to repeat the process on the other side

Answer 9

my result is different:

It's wrong @satellite73 ?

Answer 10

yes it is is wrong

Answer 11

you have 3 factors to consider, not two
they are \(x+2, x, x-3\)

Answer 12

x + 2 - - - (-2) + + + + + + + + + + + +
x - - - - - - - - 0 + + + + + +
x-3 - - - - - - - - - - - - 3 + + + +

product - - - (-2) ++ +(0) - - (3) + + +

Answer 13

Thanks @satellite73 ! The result is : [-3,-1] union [1,3] ^^