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Thread: Use of modulus..

  1. #1
    Newbie findmehere.genius's Avatar
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    Use of modulus..

    I have tried a lot of questions on Modulus...yet it creates trouble for me when it comes to what would be the answer to a questions like these:
    $\displaystyle ||x-2|-2|=x$ find value of x
    and one more, $\displaystyle |3-|2x+5||=x+2$ find $\displaystyle x$..


    answer to the first one is $\displaystyle 0<=x<=2$ and of second one is 0, -4/3

    I will be thankful to any sort of help.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by findmehere.genius View Post
    I have tried a lot of questions on Modulus...yet it creates trouble for me when it comes to what would be the answer to a questions like these:
    $\displaystyle ||x-2|-2|=x$ find value of x
    and one more, $\displaystyle |3-|2x+5||=x+2$ find $\displaystyle x$..


    answer to the first one is $\displaystyle 0<=x<=2$ and of second one is 0, -4/3

    I will be thankful to any sort of help.
    Hi

    $\displaystyle ||x-2|-2|=x$ find value of x

    Since there is |x-2| you need to consider 2 cases :
    1) $\displaystyle x\leq 2$
    In this case |x-2| = 2-x (because $\displaystyle x-2 \leq 0$)
    The equation becomes |-x|=x or |x|=x (because |-x|=|x|)
    which means $\displaystyle x\geq 0$
    The solutions in this case are therefore $\displaystyle 0 \leq x \leq 2$

    2) $\displaystyle x\geq 2$
    In this case |x-2| = x-2 (because $\displaystyle x-2 \geq 0$)
    The equation becomes |x-4|=x

    2 sub-cases :
    2a) $\displaystyle x\leq 4$
    The equation becomes 4-x=x which means x=2
    2b) $\displaystyle x\geq 4$
    The equation becomes x-4=x which has no solution
    The solutions in this case are therefore $\displaystyle x = 2$

    The overall solutions are therefore $\displaystyle 0 \leq x \leq 2$
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  3. #3
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    The method is the same for the second one

    $\displaystyle |3-|2x+5||=x+2$ find $\displaystyle x$

    Case 1 : $\displaystyle x \leq -\frac52$
    The equation becomes $\displaystyle |3-(-2x-5)|=x+2$ or $\displaystyle |2x+8|=x+2$

    Case 1a : $\displaystyle x \leq -4$
    The equation becomes $\displaystyle -(2x+8)=x+2$ or $\displaystyle x = -\frac{10}{3}$
    But $\displaystyle -\frac{10}{3} > -4$ therefore there is no solution for this case

    Case 1b : $\displaystyle x \geq -4$
    The equation becomes $\displaystyle 2x+8=x+2$ or $\displaystyle x = -6$
    But $\displaystyle -6 < -4$ therefore there is no solution for this case

    Case 2 : $\displaystyle x \geq -\frac52$
    The equation becomes $\displaystyle |3-(2x+5)|=x+2$ or $\displaystyle |-2x-2|=x+2$

    Case 2a : $\displaystyle x \leq -1$
    The equation becomes $\displaystyle -2x-2=x+2$ or $\displaystyle x = -\frac43$
    And $\displaystyle -\frac43 \leq -1$ therefore $\displaystyle -\frac43$ is one solution

    Case 2b : $\displaystyle x \geq -1$
    The equation becomes $\displaystyle 2x+2=x+2$ or $\displaystyle x = 0$
    And $\displaystyle 0 \geq -1$ therefore $\displaystyle 0$ is one solution

    Finally the set of solutions is -4/3 and 0
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  4. #4
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    Hello, findmehere.genius!

    I graphed the first one . . . Got the same answer as running-gag.


    $\displaystyle \bigg||x-2|-2\bigg|\:=\:x$ . Find value of $\displaystyle x.$
    I'll graph it in baby steps . . .

    We know what $\displaystyle y \:=\:x$ looks like:
    Code:
                    |
                    |     *
                    |   *
                    | *
          - - - - - * - - - - -
                  * | 
                *   |
              *     |
                    |

    Then we have: .$\displaystyle y \:=\:|x|$
    The absolute value says: anything below the x-axis is reflected upward.
    Code:
                    |
            *       |       *
              *     |     *
                *   |   *
                  * | *
          - - - - - * - - - - -
                    |
                    |

    Then $\displaystyle y \:=\:|x-2|$ moves the graph 2 units to the right.
    Code:
            *   |           *
              * |         *
                *       *
                | *   *
          - - - + - * - - - - -
                |   2
                |

    Then $\displaystyle y \:=\:|x-2| -2$ moves the graph down 2 units.
    Code:
                |
          *     |             *
            *   |           *
              * |         *
          - - - * - - - * - - -
                | *   *
                |   *
                |

    Finally, $\displaystyle y \:=\:||x-2| - 2|$
    . . Anything below the $\displaystyle x$-axis is reflected upward.
    Code:
                |
          *     |             *
            *   |   *       *
              * | *   *   *
          - - - * - + - * - - - -
                |   2
                |
    There!


    What does this graph have in common with $\displaystyle y = x$ ?
    . . (See the first graph.)


    We find that the graphs coincide for: .$\displaystyle 0 \:\leq\:x\:\leq\:2$

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