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Solving modulus functions

  1. Mar 6, 2006 #1

    Hootenanny

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    I need to solve:
    [tex]\left| 2x-3 \right| = 5 - x^2[/tex]
    I started by squaring the equation because modulus functions can only be positive and obtained:
    [tex]x^4 -14x^2 +12x +1 = 0[/tex]
    I haven't olved any quadnomial equations before, so I don't know where to start. Any help would be much appreciated.
     
  2. jcsd
  3. Mar 6, 2006 #2

    VietDao29

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    Nah squaring both sides just turns your little problem into a monster...
    It'll help if you note that:
    |A| >= 0.
    So if you are saying that |A| = B, then is it obvious that B should be non-negative, too?
    Now if B is non-negative, and we have |A| = B, so that means:
    A = B, if A >= 0
    A = -B, if A < 0, right?
    So all you have to do is to solve the system of equations:
    [tex]\left\{ \begin{array}{l} 5 - x ^ 2 \geq 0 \\ \left[ \begin{array}{l} 2x - 3 = 5 - x ^ 2 \\ 2x - 3 = x ^ 2 - 5 \end{array} \right. \end{array} \right.[/tex]
    Can you get it? :)
     
  4. Mar 6, 2006 #3

    Hootenanny

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    I get:
    [tex]x\leq\sqrt{5}[/tex]
    Then for the first equation [itex]x=2[/itex] and [itex]x=-4[/itex].
    And for the second; [itex]2\pm 2\sqrt{3}[/itex] but we must ignore the [itex]2 + 2\sqrt{3}[/itex] because it lies outside the inequality.
    Ive got three answers what have I done wrong?
     
  5. Mar 6, 2006 #4

    shmoe

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    The split to the cases 2x-3=5-x^2 and 2x-3=x^2-5 had additional restrictions on x that came from the absolute value sign, you haven't taken this into account yet.
     
  6. Mar 6, 2006 #5

    VietDao29

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    This is wrong: What if x = -7, -7 < sqrt(5), but 5 - (-7)2 = 5 - 49 = -44 < 0!
    What you should get is:
    [tex]-\sqrt{5} \leq x \leq \sqrt{5}[/tex]
    This is correct. :)
    However, -4 is not a valid solution, it's outside the range.
    Nope, you've solved the second equation incorrectly. You forget to devide it by 2a (i.e: 2).
    Can you go from here? :)
    ------------
    By the way, one can always check their answer by plugging the solution back to the equation. For example: x = 2 is one of the solution. So:
    |2x - 3| = |2 . 2 - 3| = 1
    5 - x2 = 5 - 22 = 1.
    And hurray!!! 1 = 1.
    Can you get this? :)
     
    Last edited: Mar 6, 2006
  7. Mar 6, 2006 #6

    VietDao29

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    Nah, we just need the "restriction" 5 - x2 >= 0. If: 2x - 3 = 5 - x2, then it's obvious that 2x - 3 >= 0.
    And if: 2x - 3 = -(5 - x2) = x2 - 5, then it's obvious that 2x - 3 <= 0. No?
    So one is enough, I think. :wink:
     
  8. Mar 6, 2006 #7

    Hootenanny

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    Ahh yes, I remember the critical values now. It's along time since I've practised solving inequalities.

    So
    [tex]x = \frac{2\pm 2\sqrt{3}}{2} \Rightarrow x = \pm\sqrt{3}[/tex]
    Both solutions lie with the inequality so the solutions are [itex] x=-4, x=-\sqrt{3},x = \sqrt{3}, x=2[/itex]

    Does that look ok?
     
  9. Mar 6, 2006 #8

    VietDao29

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    Nah, this is again wrong... :tongue2:
    It should read:
    [tex]x = 1 \pm \sqrt{3}[/tex] :)
    No, that does not, you may want to re-check that, there are up to 2 solutions that do not satisfy the inequality 5 - x2 >= 0.
    You'll have only 2 valid solutions left.
    Can you go from here? :)
     
  10. Mar 6, 2006 #9

    Hootenanny

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    O dear, does the lack of sleep show?

    So the only valid solutions are:
    [tex]x=2[/tex]
    [tex]x=1-\sqrt{3}[/tex]

    Thank's very much for your help.
     
  11. Mar 6, 2006 #10

    VietDao29

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    Pretty much. :approve:
    Yes, this is correct.
    Congratulations, :)
    It's my pleasure. :smile:
     
  12. May 20, 2010 #11
    can i have solution of the Question if f(x) = 1-x/1+x,x>0 then f(f(x)) +f(f(1/x))
     
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