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Thread: [SOLVED] Solution Set of another expression?

  1. #1
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    [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?
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  2. #2
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    Quote Originally Posted by mark1950 View Post
    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
    Last edited by Rapha; Jun 15th 2009 at 05:34 AM.
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    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.
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  4. #4
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    Quote Originally Posted by mark1950 View Post
    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!
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  5. #5
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    Quote Originally Posted by mark1950 View Post
    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
    $
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  6. #6
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    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!
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  7. #7
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    hi
    $\displaystyle \frac{{x}^{2}-3\,x-4}{x} > 0 $ $\displaystyle (x \neq 0)$
    solve $\displaystyle {x}^{2}-3\,x-4$
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