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Absolute value with quadratic

  1. Jun 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Very embarrassing that I had problems with this. I managed to solved it at the end, but I am posting this for two reasons

    1) I like someone else to work it out in a different method, I wnt to see what other approaches there are

    2) Whether mine is taking too long or not. I mean I solved it, but any unnecessary steps?

    3) Check whether I am right or wrong lol

    [tex]|x^2 + 6x + 16| < 8[/tex]

    3. The attempt at a solution

    At first I was going to do

    [tex] - 8 < x^2 + 6x + 16 < 8[/tex]

    Then I realize it was hopeless

    So then I did

    [tex] (x^2 + 6x + 16) < 8[/tex] and [tex] - ( x^2 + 6x + 16 ) < 8[/tex]


    (1) [tex] (x^2 + 6x + 16) < 8[/tex]


    [tex] x^2 + 6x + 8 < 0[/tex]


    [tex] (x+4)(x+2) < 0[/tex]

    Did some test points and found that [tex] x \in (-4,-2) [/tex] is a solution

    (2) [tex] -(x^2 + 6x + 16) < 8[/tex]

    [tex] -(x+4)(x+2) < 0[/tex]

    [tex](x+4)(x+2) > 0[/tex]

    Now here is the problem should i have even divide that -1 and switch the inequality signs? I could and I would get some meaningless answer like [tex]x \in (-\infty,-4)[/tex]

    Anyways I threw it back in

    [tex](-x-4)(x+2) > 0[/tex]

    x is still 4, so no change, neither did the solution

    So my solution remains as [tex]x \in (-4,-2)[/tex]
     
  2. jcsd
  3. Jun 25, 2011 #2

    ehild

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    The second line is wrong.


    ehild
     
  4. Jun 25, 2011 #3

    Mentallic

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    This isn't hopeless. Solve both cases separately, mainly:

    [tex]-8<x^2+6x+16[/tex]
    [tex]x^2+6x+16<8[/tex]
    and then find the intersection of each solution set.

    This is equivalent to

    And you can improve on this part

    By simply drawing a rough sketch of the quadratic (you should be able to visualize it in your head even).

    And ehild has pointed out your error.
    What you did is basically say that since a<b (from your first inequality) then a-b<0, but if -a<b, then -a-b<0 but -a-b is not equal to -(a-b) as you did.

    One more point: Even though you solved the wrong quadratic,

    This would not be true since you need to find the intersection of the solution sets [itex]x\in(-\infty,-4)[/itex] and [itex]x\in(-4,-2)[/itex] which is the empty set (x cannot be in both at the same time, so there is no x satisfying these solution sets).
     
  5. Jun 25, 2011 #4
    But I thought

    |x| < a

    is

    -x<a and x <a

    A definition of piecewise?
     
  6. Jun 25, 2011 #5

    Mentallic

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    Look more closely, that's not what the problem was.
     
  7. Jun 25, 2011 #6

    Mark44

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    Or equivalently, x > -a and x < a. This is normally written as -a < x < a, assuming a is a positive number.
    ???
    I don't understand your thinking here.
     
  8. Jun 25, 2011 #7

    SammyS

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    To take what ehild wrote, one or two steps farther;

    If [itex]- ( x^2 + 6x + 16 ) < 8\,, [/itex] then [itex]-(x^2+6x+24)<0\,.[/itex]

    Now solve that.
     
  9. Jun 25, 2011 #8

    SammyS

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    You might try completing the square for the quadratic, x2 + 6x + 16 .

    x2 + 6x + 16 = x2 + 6x + 9 + 7
    = (x + 3)2 + 7​

    For on thing, this tells you that the minimum value for the quadratic is 7.
     
  10. Jun 26, 2011 #9
    But there aer no real solutions for x^2 + 6x + 24
     
  11. Jun 26, 2011 #10

    ehild

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    That means, it never crosses the x axis. It is either positive or negative everywhere. Which one?

    ehild
     
  12. Jun 26, 2011 #11

    Mentallic

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    Notice the post just before yours:

     
  13. Jun 26, 2011 #12
    Yeah so you can't solve it, it's always positive and never less than 0.
     
  14. Jun 26, 2011 #13

    Mentallic

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    If we're trying to solve [itex]-(x^2+6x+24)<0[/itex] this is equivalent to solving [itex]x^2+6x+24>0[/itex] :wink:
     
  15. Jun 26, 2011 #14

    ehild

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    So −8<x2+6x+16 is always true.

    ehild
     
  16. Jun 26, 2011 #15
    Oh okay so that means all values of x will work, but doesn't this contradict my answer that (-4,-2)
     
  17. Jun 26, 2011 #16

    ehild

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    No, the other side of the inequality holds only in the interval (-4,-2)

    ehild
     
  18. Jun 26, 2011 #17

    Mentallic

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    x has to be satisfied by both solution sets at the same time. It's called taking the intersection.
     
  19. Jun 26, 2011 #18

    SammyS

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    Well, [itex]x^2+6x+16[/itex] is positive for all x.

    What does that tell you about [itex]\left|x^2+6x+16\right|\,?[/itex]
     
  20. Jun 26, 2011 #19
    Not changing...
     
  21. Jun 26, 2011 #20

    rock.freak667

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    So if |x2+6x+16| means to take the positive value of "x2+6x+16" and "x2+6x+16" is always positive, then you do you really need the modulus sign?
     
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