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Number of roots

  1. May 4, 2008 #1
    how do i find the number of roots for a curve that has dy/dx

    2x^4 -20x^2 + 50.

    if i substitute y=x^2 and use the discriminant formula i get

    b^2 - 4ac = 400 - 4 x 2 x 50
    = 400 - 400
    = 0

    This way says there 1 root, answers say it has 2. Which method am i meant to use for this?

    if i factorise i get 2(y-5)^2

    which is also 1?
     
    Last edited: May 4, 2008
  2. jcsd
  3. May 4, 2008 #2

    malty

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    Do you know of a way to find how many times the curve crosses the x-axis?

    Think of info needed to sketch this curve. .
     
    Last edited: May 4, 2008
  4. May 4, 2008 #3
    yes i know 2 ways... factorizing and discriminant. and they both say 1 root..
     
  5. May 4, 2008 #4

    malty

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    Are you familar with critical points (Maxima, Minima, points of inflection . .) of a curve?

    It looks to me like you are only finding the roots of the tangent to the curve?
     
  6. May 4, 2008 #5

    HallsofIvy

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    Are you look for the zeroes of y= 2x^4 -20x^2 + 50 or y such that dy/dx= 2x^4 -20x^2 + 50?

    Obviously, 2x^4 -20x^2 + 50= 2(x^4- 10x+ 25)= 2(x^2- 5)(x^2+ 5) has two real roots- they are [itex]\pm\sqrt{5}[/itex].

    But if you mean y such that dy/dx= 2x^4 -20x^2 + 50, there is no way of telling. You lose an additive constant when you differentiate y and how many times y is 0 depends on that constant.
     
  7. May 4, 2008 #6
    2x^4 -20x^2 + 50= 2(x^4- 10x+ 25)= 2(x^2- 5)(x^2+ 5) Is wrong.?

    2(x^2- 5)(x^2+ 5) = 2(X^4 -25) not 2(x^4 - 10x + 25)
     
  8. May 4, 2008 #7

    epenguin

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    It should be easy to sketch what a curve of dy/dx gainst x and y against x when dy/dx is what you said (in the form you factorised it). The height of the second of these is not determined, as mentioned, unless that was given too in your original problem. But basically you should then see there is only one general possibility, plus one special case.
     
  9. May 4, 2008 #8

    HallsofIvy

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    Ooops! That's embarrassing! 2x^4 -20x^2 + 50= 2(x^4- 10x+ 25)= 2(x^2- 5)^2 which has 2 distinct roots, each a double root.

    Thanks, DeanBH.

    I'm still wondering what the original problem really was!
     
  10. May 4, 2008 #9
    it still makes no sense.

    it says find stationary points on the curve blahblahblah.
    the point is the curve has dy/dx 2x^4 -20x^2 + 50.

    how the hell do i go about finding it has 2.

    it looks like it has 1.
     
  11. May 4, 2008 #10

    Hootenanny

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    Ahh, the question make sense now!

    Why do you think that there is only one root? Both you and Halls have shown that it has two roots and therefore two stationary points.
     
  12. May 4, 2008 #11
    Read Hall's post!

    [tex]2(x^2-5)^2=0[/tex]

    Can you find those 2 roots?
     
  13. May 4, 2008 #12
    that looks like one root to me, why is that 2.
     
  14. May 4, 2008 #13

    Hootenanny

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    Cancel the two and take the square root of both sides. Does that make it any easier?
     
  15. May 4, 2008 #14
    no.

    i dont know what you are talking about
     
  16. May 4, 2008 #15
    Solve ...

    [tex]x^2-5=0[/tex]

    What do you get?
     
  17. May 4, 2008 #16
    oooooooooooooooooooohhhhhhhhhhhhh craaaaaaaaaaaaaaaappppppppppp>

    i was ignoring the fact it was X^2. taking it as X
    me being retarded
     
  18. May 4, 2008 #17
    Yeah I was getting worried for a sec :p hehe, don't worry about it! We all have brain farts, just pray it isn't during an exam!!!
     
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