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inkyvoyd

  • 3 years ago

How does one solve a polynomial absolute value inequality? say, |(x-a)(x-b)(x-c)|>d

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  1. inkyvoyd
    • 3 years ago
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    No joke, I got a prob on a quiz for my AP BC calc class and I miserably got it wrong...

  2. inkyvoyd
    • 3 years ago
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    the lesson mentioned inequalities for polynomials, but not polynomial abs value probs.

  3. inkyvoyd
    • 3 years ago
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    And, if I'm supposed to use a graphing calculator, tell me, because I honestly have no idea.

  4. phi
    • 3 years ago
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    I think these are tricky. I would re-write as (x-a)(x-b)(x-c) - d > 0 or -(x-a)(x-b)(x-c) - d > 0 multiply out and then factor to find the zeros then look for intervals where the product of the factors are positive very messy...

  5. inkyvoyd
    • 3 years ago
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    Yes, but linear absolute value problems can have extraneous solutions...

  6. phi
    • 3 years ago
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    then check the possible answers in the original equation

  7. inkyvoyd
    • 3 years ago
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    So I probs just want a graphing calculator?

  8. phi
    • 3 years ago
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    off hand, I don't know. do you have a specific example?

  9. inkyvoyd
    • 3 years ago
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    Let me access my course. Sec

  10. inkyvoyd
    • 3 years ago
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    |x^3-x^2+x-1|<3 "Round your answer to three decimal places and place in ascending order with a comma in between." Answer: -0.811<x<-0.732 Incorrect Response (-.811, 1.743, -.811,1.743, -0.811,1.743, -0.811, 1.743)

  11. inkyvoyd
    • 3 years ago
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    yeah, and my answer was from wolfram. I have no idea what I was supposed to do. and the lesson was not helpful at all. Didn't even mention it.

  12. phi
    • 3 years ago
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    I'll think about it, as this is not something I've looked at very carefully. But I not that x=1 gives 0 so 1 is part of the solution space. Maybe by tomorrow I'll have something to post.

  13. inkyvoyd
    • 3 years ago
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    Alright. Thanks for looking at it with me.

  14. amistre64
    • 3 years ago
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    if you graph the inner polynomial, whenever it goes negative, it just mirrors above the x axis

  15. amistre64
    • 3 years ago
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    well, the curve does, then pick a point that is clearly not on the curve to test

  16. amistre64
    • 3 years ago
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    |dw:1348175550151:dw|

  17. amistre64
    • 3 years ago
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    |dw:1348175643801:dw|

  18. inkyvoyd
    • 3 years ago
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    o.o

  19. inkyvoyd
    • 3 years ago
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    it's a fake 6 degree polynomial!

  20. inkyvoyd
    • 3 years ago
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    |dw:1348176111960:dw|

  21. inkyvoyd
    • 3 years ago
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    Mmm.

  22. amistre64
    • 3 years ago
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    its just the normal 3 degree with its negative parts popped up; since the absolute value of a number is never negative.

  23. inkyvoyd
    • 3 years ago
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    Yeah. and then we draw that line. any way of an analytic solution, or are we just going to have to graph this guy?

  24. amistre64
    • 3 years ago
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    im curious why the y=3 line

  25. amistre64
    • 3 years ago
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    i see it now

  26. amistre64
    • 3 years ago
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    yes, that y=3 line represents where the x axis would be if we subtract 3 from each side

  27. amistre64
    • 3 years ago
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    |x^3-x^2+x-1| < 3 consider the values such that (x^3-x^2+x-1) = 3 -(x^3-x^2+x-1) = 3

  28. amistre64
    • 3 years ago
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    x^3-x^2+x-4 = 0 and x^3-x^2+x+2 = 0 there is no good way to do a general cubic solution is there

  29. inkyvoyd
    • 3 years ago
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    yeah I'm not cardano, I have 30 minutes to do this xD. I think it was a calculator prblem lol

  30. amistre64
    • 3 years ago
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    do you need exact or approx ?

  31. inkyvoyd
    • 3 years ago
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    approx.

  32. amistre64
    • 3 years ago
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    x = 1.743 and -0.811 is what the google gives me as approx zeros

  33. amistre64
    • 3 years ago
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    we can test those in the original format

  34. inkyvoyd
    • 3 years ago
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    kay

  35. amistre64
    • 3 years ago
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    at x=1.743; |x^3-x^2+x-1| -3 = 0.000xxx at x=-0.811; |x^3-x^2+x-1| -3 = 0.00xxxx and of course, when x=0, 1-3 is less than 0 so it should be an interval like: (-0.811,1.743) or however exact you need it to be

  36. inkyvoyd
    • 3 years ago
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    I see. So that's what I'm supposed to do. Thank you.

  37. amistre64
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=%7Cx%5E3-x%5E2%2Bx-1%7C+-3%3C0 it even has the interval as a number line at the bottom

  38. phi
    • 3 years ago
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    Here is what I came up with |x^3-x^2+x-1|<3 we can do this with a graphing calculator Case 1: x^3-x^2+x-1 < 3 x^3-x^2+x-4 < 0 Examine the roots of x^3-x^2+x-4 = 0 Find 1 real root at x= 1.743 Graph of the function shows f(x)<0 for x < 1.743 Case 2: -(x^3-x^2+x-1) < 3 or, negating and flipping the sense of the relational operator x^3-x^2+x-1 > -3 x^3-x^2+x+2 > 0 Find the roots of f(x)= x^3-x^2+x+2 =0 1 real root at x= -0.811 Graph shows f(x) > 0 for x > -0.811 Intersection of both regions x < 1.743 x > -0.811 Gives -0.811 < x < 1.743

  39. inkyvoyd
    • 3 years ago
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    I get why wolfram gave me the wrong answer. Cause it split it up into two inequalities. So mad lol. and @phi , thanks for the answer :D

  40. phi
    • 3 years ago
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    Sounds like you are ready for these... give 'em hell

  41. inkyvoyd
    • 3 years ago
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    Too late :o I got into BC calc 2.5 weeks late and I'm doing catch up work. that right there was week 1 or something, and I'm at week 2.5 tomorrow ;) But thankfully I understand the concept, so I can take it on in event of unit test.

  42. inkyvoyd
    • 3 years ago
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    It's not even calculus and I got it wrong. xS

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