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Math Help - Inequalities.

  1. #1
    Super Member Showcase_22's Avatar
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    Inequalities.

    I have two inequality questions that I need help solving:



    As you can see, i've ended up with all the different inequalities but when I combine then I get the wrong answer.

    help me please!
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  2. #2
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    Hello, Showcase_22!

    Here's the first one.
    . . Your punchlines are a bit off . . .


    x - 4 \;< \;x(x-4) \;<\;5

    The left side: . x - 4 \:<\:x(x-4)

    We have: . x - 4 \:<\:x^2-4x \quad\Rightarrow\quad 0 \:< \:x^2-5x + 4 \quad\Rightarrow\quad x^2-5x + 4 \:>\:0

    Factor: . (x - 1)(x - 4) \:>\:0


    There are two cases: [1] both factors are positive, [2] both factors are negative.

    . . [1]\;\;\begin{Bmatrix}x - 1 &> &0 & \Rightarrow & x & > & 1 \\ x - 4 & > & 0 & \Rightarrow & x & > & 4 \end{Bmatrix} \quad\Rightarrow\quad x \:>\:4

    . . [2]\;\;\begin{Bmatrix}x - 1 &<&0 & \Rightarrow & x &<& 1 \\ x - 4 &<& 0 & \Rightarrow & x &<& 4 \end{Bmatrix} \quad\Rightarrow\quad x \:<\: 1

    Therefore: . \boxed{x < 1\;\text{ or }\;x > 4}



    The right side: . x(x-4) \:<\:5 \quad\Rightarrow\quad x^2 - 4x - 5 \:<\:0

    Factor: . (x+1)(x-5) \:<\:0


    There are two cases: [1] the factors are (-)(+), [2] the factors are (+)(-).

    [1]\;\;\begin{Bmatrix}x + 1 &<& 0 & \Rightarrow & x &<& \text{-}1 \\<br />
x - 5 &>& 0 & \Rightarrow & x &>& 5 \end{Bmatrix}\quad\Rightarrow\quad \text{impossible}

    [2]\;\;\begin{Bmatrix}x + 1 &>& 0 & \Rightarrow & x &>& \text{-}1 \\<br />
x-5 &<&0 & \Rightarrow & x &<& 5 \end{Bmatrix} \quad\Rightarrow\quad \boxed{\text{-}1 \:< \:x \:<\:5}



    To determine the final solution, I had to make pretty pictures . . .
    Code:
    
                  x < 1  or  x > 4
    
            = = = = o - - - - - o = = = =
                    1           4
    
    
                      -1 < x < 5
    
          - - o = = = = = = = = = = = o - -
             -1                       5

    Combining them, we have:
    Code:
    
          - - o = = o - - - - - o = = o - -
             -1     1           4     5

    Therefore: . \boxed{\;(\text{-}1 \:<\: x\:<\:1)\;\text{ or }\;(4 \:<\:x\:<\:5)\;}

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  3. #3
    Moo
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    Hi

    For the second one... ouch !

    You have \frac{|x|-3}{-x^2-1} \ge 1

    Note that -x^2-1=-(x^2+1). Therefore, this is always negative !

    And remember, when you multiply an inequality by a negative number, its direction is changed. This means that if you multiply both sides by -x^2-1, you'll have :

    |x|-3 ~{\color{red} \le} -(x^2+1)

    Now, see the two situations |x|=-x (if x<0) and |x|=x (if x>0)
    If you're not sure, you can do the special case x=0...
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  4. #4
    Super Member Showcase_22's Avatar
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    Thanks Soroban and Moo. I get what I was doing wrong and thanks!

    I was comparing my worng answers with the worked solutions I had on another piece of paper. I couldn't follow their method since they were doing some odd stuff with moduli (not that odd now i'm comparing it with correct answers ).
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