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amorfide Group Title

Soooooo I need help with a modulus/absolute value equation

  • 2 years ago
  • 2 years ago

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  1. amorfide Group Title
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    |dw:1350320124515:dw|

    • 2 years ago
  2. amorfide Group Title
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    any method just explain

    • 2 years ago
  3. amorfide Group Title
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    @ParthKohli

    • 2 years ago
  4. amorfide Group Title
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    |dw:1350320261447:dw| my graph

    • 2 years ago
  5. zugzwang Group Title
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    If |x| = y, then x = y OR x = -y

    • 2 years ago
  6. amorfide Group Title
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    yes i get that part, so this gives |dw:1350321371149:dw|

    • 2 years ago
  7. amorfide Group Title
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    then figure out x in both cases but i will get 4 answers, it only crosses 3 times, how do i know which one to eliminate?

    • 2 years ago
  8. 2le Group Title
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    |dw:1350321424876:dw| what you drew looks more exponential. The graph for y=|4/x| is linear with a jump at x=0.

    • 2 years ago
  9. zugzwang Group Title
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    There must be one for which x/2 + 3 is negative. That's an impossibility.

    • 2 years ago
  10. zugzwang Group Title
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    Because x/2 + 3 is the absolute value of something...?

    • 2 years ago
  11. amorfide Group Title
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    |dw:1350321988874:dw| |dw:1350322071606:dw|

    • 2 years ago
  12. amorfide Group Title
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    so what would my final answers be

    • 2 years ago
  13. zugzwang Group Title
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    Are you sure about the first part?

    • 2 years ago
  14. zugzwang Group Title
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    I think it should be -3 (+ or -) sqrt (21)

    • 2 years ago
  15. zugzwang Group Title
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    oh wait, sorry... my bad

    • 2 years ago
  16. zugzwang Group Title
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    in any case, -3-sqrt(17), try substituting that for x in x/2 + 3

    • 2 years ago
  17. zugzwang Group Title
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    And remember -3-sqrt(17) < -3-sqrt(16) = -7

    • 2 years ago
  18. amorfide Group Title
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    thank you!

    • 2 years ago
  19. zugzwang Group Title
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    No problem... Sorry about my pathetic arithmetic :P

    • 2 years ago
  20. amorfide Group Title
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    still made sense in a way :P

    • 2 years ago
  21. ParthKohli Group Title
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    \[\rm \left|{4 \over x}\right| = {x \over 2} + 3 \qquad \Rightarrow \qquad {4 \over x} = \pm\left({x \over 2 }+3 \right)\]Let's solve both.\[\rm {4 \over x}={x \over 2}+3\]so\[8 = \rm x^2 + 6x\]therefore\[\rm x^2 + 6x-8=0 \]

    • 2 years ago
  22. ParthKohli Group Title
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    For the other solution,\[{4 \over \rm x } = \rm -{x \over 2}-3\]so\[8=-\rm x^2 - 6x\]therefore\[\rm x^2 +6x +8=0\]

    • 2 years ago
  23. terenzreignz Group Title
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    And as @zugzwang was kind enough to point out, make sure all values of x lead to \[\frac{x}{2}+3\] being nonnegative.

    • 2 years ago
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