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Equation: Absolute value

  1. Jan 13, 2014 #1
    Solve |x^2-5|=4|x|.


    I tried to rewrite it as:

    (sqrt(x^2-5))^2=4*(sqrt(x))^2

    Is this the right way to solve the equation?
     
  2. jcsd
  3. Jan 13, 2014 #2

    SammyS

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    No.

    For one thing, each side of the original expression exists for all values of x.

    For your equation, (sqrt(x^2-5))^2=4*(sqrt(x))^2 , :

    The left hand side exists only if x ≥ √(5) or x ≤ -√(5).

    The right hand side exists only if x ≥ 0 .
     
  4. Jan 13, 2014 #3

    haruspex

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    Did you mean to apply those operations in the other order: square then square root?
    That would work, but why bother square-rooting?
     
  5. Jan 13, 2014 #4
    I make a new attempt.

    I put the equation in two possible cases.

    Case 1: x^2-5=4x
    x^2-4x-5=0
    x=5 and x=-1

    Case 2: x^2-5=4(-x)
    x^2+4x-5=0
    x=-5 and x=1
     
  6. Jan 13, 2014 #5

    Mark44

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    That should be x = 5 OR x = -1.
    Also, you're omitting something important. In Case 1, your assumption (not stated) is that x ≥ 0. How does this assumption fit in with your solutions?
    Same here - it should be x = -5 OR x = 1. The assumption in this case is that x < 0. How does this assumption fit in with the solutions from this case?
     
  7. Jan 13, 2014 #6
    Of course, I meant "or" not "and"

    It's just x=5 that fit in case 1
    In case 2 it's only x=-5 that fit.

    How do you make the assumptions?
     
  8. Jan 13, 2014 #7

    Ray Vickson

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    Since absolute-value equations can be tricky, the first step in such cases should be to plot both sides; that is, draw graphs of y = |x^2-5| and y = 4|x|, to see where they cross. That will help you to keep things straight. Be sure to make the x-range wide enough to show all the possibilities.
     
  9. Jan 13, 2014 #8

    Mark44

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    When you change |x| to x, you are assuming that x ≥ 0. When you change |x| to -x, you are assuming that x < 0.

    One other thing is that when you change |x2 - 5| to x2 - 5, you are assuming that x2 - 5 ≥ 0, or equivalently, that x ≥ √5 or that x < -√5.
     
  10. Jan 13, 2014 #9
    They cross at x=-5, x=-1, x=1, x=5.

    Now I'm lost. The x^2 term makes it harder to get it right.

    Could someone show how this should be solved or make an exemple. I can't find any similar task with the x^2 term in it. I would be very grateful.
     
  11. Jan 13, 2014 #10
    I usually write it as three intevals, but with this I can't do that.
     
  12. Jan 13, 2014 #11

    Mark44

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    Make four cases (hence four intervals), since you have four different possibilities.

    1. x ≥ √5
    2. 0 ≤ x < √5
    3. -√5 ≤ x < 0
    4. x < -√5

    Case 1. x ≥ √5
    For the first case, since x ≥ √5, we can say with certainty that x > 0.
    Then the original equation simplifies to
    x2 - 5 = 4x
    ==> x2 - 4x = 5 = 0
    ==> x = 5 or x = -1
    Since our assumption was that x ≥ √5, we discard x = -1.
    Solution for case 1: x = 5

    Now look at the other three cases.
     
  13. Jan 13, 2014 #12
    Case 2: 0 ≤ x < √5

    x has to be positive. 0 or more, but less then √5

    (here I'm not sure how to use the signs)

    Should it be
    -(x^2-5)=4x
    or
    (x^2-5)=-4x
    Both gives x=-5 and x=1
     
  14. Jan 13, 2014 #13
    I did it like this:

    Case 2:
    -(x^2-5)=4x
    Solution: x=1

    Case 3:
    -(x^2-5)=-4x
    Solution: x=-1

    Case 4:
    -(x^2-5)=4x
    Solution: x=-5
     
  15. Jan 13, 2014 #14

    haruspex

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    You can avoid all this breaking into cases. Just square both sides of the original equation to get rid of the modulus signs. This gives you a quadratic in x2. Solve. Final step is to check you have not introduced any extra solutions.
     
  16. Jan 14, 2014 #15

    ehild

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    Yes, [tex]|A|=\sqrt{A^2}[/tex]
    so
    [tex]\sqrt{(x^2-5)^2}=4\sqrt{x^2}[/tex]

    eliminate the square root by squaring both sides of the equation, and solve as haruspex suggested.


    ehild
     
  17. Jan 14, 2014 #16
    Okay, if I do so I'll get:

    x^4-26x^2+25=0

    and when I type it in Wolfram I get the solution:
    x=1
    x=-1
     
  18. Jan 14, 2014 #17
    I made a mistake when I wrote it in Wolfram.

    Now I got it right!
     
  19. Jan 14, 2014 #18

    Mentallic

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    You don't need a calculator to figure this out. Let [itex]u=x^2[/itex], then you'll have a quadratic in u that can be factorized. Once you have two solutions for u, solve for x.
     
  20. Jan 14, 2014 #19
    Yes, it's easy, I was outside with no pen or paper so I took some help from Wolfram.
    Math becomes hard when your in a hurry.
     
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