# [SOLVED] Solution Set of another expression?

• Jun 15th 2009, 05:14 AM
mark1950
[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 AM
Rapha
Quote:

Originally Posted by mark1950
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?

{-1 < x < 0 or x > 4 } is the right solution
show us your steps, then we are going to find your mistake.

Yours
Rapha
• Jun 15th 2009, 05:35 AM
mark1950
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 AM
stapel
Quote:

Originally Posted by mark1950
This is how I did it.

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

By multiplying through by "x" and not flipping the inequality sign, you were assuming that x was positive! (Wink)
• Jun 15th 2009, 05:40 AM
Quote:

Originally Posted by mark1950
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.

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 AM
mark1950
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 AM
Raoh
hi(Hi)
$\displaystyle \frac{{x}^{2}-3\,x-4}{x} > 0$ $\displaystyle (x \neq 0)$
solve $\displaystyle {x}^{2}-3\,x-4$