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Math Help - hmmm... just wondering

  1. #16
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack
    What do you get if you substitute this back into the original equations?
    Is there a x which with this as a value of y solves the equations?

    RonL

    (when I do this I get no consistent solution for x, but I could be doing the
    arithmetic wrongly )
    Using the graphical method I find an initial estimate for the solution
    is x=2.36, y=1.64, and there is probably but one real solution.

    RonL
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  2. #17
    dan
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    hmmmm.....
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  3. #18
    dan
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    Quote Originally Posted by ThePerfectHacker
    (y^2 - 6)^2 = (10+y^3)^3
    Thus,
    y^9+30y^6-y^4+300y^3+12y^2+9964=0
    is this correct...for some reason it's not getting me a good value for y...

    i'm getting tired of this problem...
    dan
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  4. #19
    dan
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    Quote Originally Posted by CaptainBlack
    Using the graphical method I find an initial estimate for the solution
    is x=2.36, y=1.64, and there is probably but one real solution.

    RonL
    what is the graphical method ??

    dan
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  5. #20
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    Quote Originally Posted by dan
    what is the graphical method ??

    dan
    Plot a graph of the curve represented by:

    <br />
x^2 +y^3 =10<br />
,

    and on the same axes plot the graph the curve represented by:

    <br />
x^3 -y^4 =-6<br />
.

    The points where these curves cross are roots of the system.

    RonL
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  6. #21
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    Quote Originally Posted by dan
    what is the graphical method ??

    dan
    You draw the graw and see where it interesects the x-axis.
    There is only one such place.
    Thus, there is only no real root.
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  7. #22
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    Quote Originally Posted by ThePerfectHacker
    You draw the graw and see where it interesects the x-axis.
    There is only one such place.
    Thus, there is only no real root.
    Try the graphs of the original two curves, which do seem to have a root.

    RonL
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  8. #23
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    Hier
    Attached Thumbnails Attached Thumbnails hmmm... just wondering-picture4.gif  
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  9. #24
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    Hier
    That seems more accurate than my initial sketch, but now we have an
    estimate we can replot about the solution.

    RonL
    Attached Thumbnails Attached Thumbnails hmmm... just wondering-gash.jpg  
    Last edited by CaptainBlack; September 7th 2006 at 09:17 AM.
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  10. #25
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    Quote Originally Posted by ThePerfectHacker
    y^4-12y^2+36=1000+300y^3+30y^6+y^9
    Thus,
    y^9+30y^6-y^4+300y^3+12y^2+9964=0
    Typo, last should be:

    y^9+30y^6-y^4+300y^3+12y^2+964=0

    RonL
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  11. #26
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    Quote Originally Posted by ThePerfectHacker
    Of course,
    x^3=y^4-6
    x=\sqrt[3]{y^4-6}
    Substitute,
    (y^4-6)^{2/3}+y^3=10
    Thus,
    (y^4-6)^{2/3}=(10+y^3)
    Typo, last should be:

    (y^4-6)^{2/3}=(10-y^3)

    This canges everything down stream along this line of solution

    RonL
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  12. #27
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    Quote Originally Posted by CaptainBlack
    Typo, last should be:

    (y^4-6)^{2/3}=(10-y^3)

    This canges everything down stream along this line of solution

    RonL
    Expanding this gives:

    <br />
y^9 + y^8 - 30 y^6 - 12 y^4 + 300 y^3 - 964=0<br />

    which may be solved numerically to give y = 1.873782741

    RonL
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  13. #28
    dan
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    Thumbs up

    Quote Originally Posted by CaptainBlack
    Expanding this gives:

    <br />
y^9 + y^8 - 30 y^6 - 12 y^4 + 300 y^3 - 964=0<br />

    which may be solved numerically to give y = 1.873782741

    RonL
    captblack, you're the man you get a +rep.
    thanks guys...case closed...untill later

    dan
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