Find the solution set of $\displaystyle x - 2 > \frac{x + 4}{x} $. The answer they gave was {x : x>4 or -1<x<0 } but my answer was {x : x>4 or x<-1}. Which is right?

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- Jun 15th 2009, 05:14 AMmark1950[SOLVED] Solution Set of another expression?
Find the solution set of $\displaystyle x - 2 > \frac{x + 4}{x} $. The answer they gave was {x : x>4 or -1<x<0 } but my answer was {x : x>4 or x<-1}. Which is right?

- Jun 15th 2009, 05:23 AMRapha
- Jun 15th 2009, 05:35 AMmark1950
This is how I did it.

$\displaystyle x - 2 > \frac{x + 4}{x} $

$\displaystyle x^2 - 2x > x + 4 $

$\displaystyle x^2 - 3x - 4 > 0 $

$\displaystyle (x - 4)(x + 1) > 0 $

Since the expression is > 0, hence, the x values are above the x -axis, right? And since the graph is a "smiling" graph, x > 4 or x < -1. That's how I got it. - Jun 15th 2009, 05:40 AMstapel
- Jun 15th 2009, 05:40 AMmathaddict

You made a mistake here . You CANNOT multiply x-2 with x because u don know whether x is positive or negative . If its positive , the sign is not affected but if its negative , the sign will be affected.

So start like this :

$\displaystyle (x-2)-\frac{x+4}{x}>0$

$\displaystyle

\frac{x(x-2)}{x}-\frac{x+4}{x}>0

$ - Jun 15th 2009, 05:55 AMmark1950
Oh! No wonder...I got the previous question wrong too. Wow. This is a new concept for me. So, I can't multiply unknowns in inequalities. Thanks, guys!

- Jun 15th 2009, 05:58 AMRaoh
hi(Hi)

$\displaystyle \frac{{x}^{2}-3\,x-4}{x} > 0 $ $\displaystyle (x \neq 0)$

solve $\displaystyle {x}^{2}-3\,x-4$