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sauravshakya Group Title

Another Fun Question Find all the real solutions: x^14=7+x^2

  • 2 years ago
  • 2 years ago

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  1. ParthKohli Group Title
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    Not sure :p

    • 2 years ago
  2. ksaimouli Group Title
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    |dw:1350910904638:dw|

    • 2 years ago
  3. ParthKohli Group Title
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    That's also the point I reached. and then \(\rm \large x^2(x^{12 }-1) - 7 = 0\)

    • 2 years ago
  4. ParthKohli Group Title
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    And then I factored it by difference of squares continuously.

    • 2 years ago
  5. ParthKohli Group Title
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    Then I reached nothing.

    • 2 years ago
  6. ksaimouli Group Title
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    |dw:1350910926540:dw|

    • 2 years ago
  7. benzilla Group Title
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    X^14 -x^2=7..........>X^2(X^12-1)...........>x^2(x-1)(x^11+x^10.......) The second expression is prime so your two real roots are zero and one

    • 2 years ago
  8. UnkleRhaukus Group Title
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    \[x^{14}=7+x^2\]\[x^{14}-x^2=7\]\[x^2(x^{12}-1)=7\]

    • 2 years ago
  9. benzilla Group Title
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    oops i made a mistake I forgot the 7

    • 2 years ago
  10. UnkleRhaukus Group Title
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    now what,

    • 2 years ago
  11. benzilla Group Title
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    The equation is then xX2(x-1)(x^11+x^10,,,,,,)=7

    • 2 years ago
  12. UnkleRhaukus Group Title
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    there are fourteen solutions

    • 2 years ago
  13. sauravshakya Group Title
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    Just find real solutions.

    • 2 years ago
  14. benzilla Group Title
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    I set this to zero which made this come out in error. This needs an iterative process to solve.

    • 2 years ago
  15. sauravshakya Group Title
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    Yep... @benzilla

    • 2 years ago
  16. mukushla Group Title
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    if there are some answers they are in the range\[\rm 1<|x|<2\]

    • 2 years ago
  17. sauravshakya Group Title
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    Yep

    • 2 years ago
  18. mukushla Group Title
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    answer will be close to 1 so letting\[x^2 \approx1\]\[x^{14}=8\]\[x=\sqrt[14]{8}\]

    • 2 years ago
  19. mukushla Group Title
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    actually\[x\approx\pm\sqrt[14]{8}\]

    • 2 years ago
  20. sauravshakya Group Title
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    What kind of solution is that???

    • 2 years ago
  21. mukushla Group Title
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    thats just a approximate solution...not exact

    • 2 years ago
  22. mukushla Group Title
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    we have\[x\approx\pm\sqrt[14]{8}=\pm1.16\]comparing this with exact solutions http://www.wolframalpha.com/input/?i=x%5E14%3D7%2Bx%5E2 its a good approaximate :)

    • 2 years ago
  23. sauravshakya Group Title
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    Yep... What about the exact solution.

    • 2 years ago
  24. sauravshakya Group Title
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    I think it is irrational

    • 2 years ago
  25. mukushla Group Title
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    well thats what engineers do...lets think like mathematicians

    • 2 years ago
  26. sauravshakya Group Title
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    Let y=x^2 then, y^7=7+y y=(7+y)^(1/7)

    • 2 years ago
  27. sauravshakya Group Title
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    |dw:1350912581733:dw|

    • 2 years ago
  28. mukushla Group Title
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    trial and error

    • 2 years ago
  29. sauravshakya Group Title
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    And that gives:|dw:1350912666885:dw|

    • 2 years ago
  30. sauravshakya Group Title
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    Thus, |dw:1350912725583:dw|

    • 2 years ago
  31. sauravshakya Group Title
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    Can I do that?

    • 2 years ago
  32. benzilla Group Title
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    Beyond a cubic function, there is not an equation like the quadratic. There are some relationships for some functions but what you have to do is yes as mukushla says trial and error. You take the number and keep plugging until you get close enough to the actual number. While we want a pure number, it will be hard to solve any other way.

    • 2 years ago
  33. ganeshie8 Group Title
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    thats very much an exact solution :)

    • 2 years ago
  34. sauravshakya Group Title
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    And @UnkleRhaukus from similar process the solution of x^3=1+x is |dw:1350913011409:dw|

    • 2 years ago
  35. UnkleRhaukus Group Title
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    hmm, good thinking

    • 2 years ago
  36. benzilla Group Title
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    If you want to try to find all factors here is a little more to warp the brain. It is all factored except for two cubic terms. X^2(x^3+i)(x^3-i) (x-1)[X-1/2-[(sqrt3)/2]i] [X-1/2+[(sqrt3)/2]i](x+1) [X+1/2+[(sqrt3)/2]i] [X-1/2-[(sqrt3)/2]i]=7 X^2(X+i)(?)(X-i)(??) (x-1)[X-1/2-[(sqrt3)/2]i] [X-1/2+[(sqrt3)/2]i](x+1) [X+1/2+[(sqrt3)/2]i] [X-1/2-[(sqrt3)/2]i]=7 The ? and ?? can be factored however i am running out of time this morning. Good luck

    • 2 years ago
  37. benzilla Group Title
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    Here it is completely factored. I couldn't resist. A lot of fun but what a mess. You will see there are a lot of roots in this equation. X^2(X+i )(X +i/2 +[sqrt3]/2)(X+i/2-[sgrt3]/2)(X-i)(X-i(1+sqrt3)/2)(X-i(1-sqrt3)/2) (x-1)[X+1/2-(sqrt3)i] [X+1/2+(sqrt3)i](x+1) [X-1/2+(sqrt3)i] [X-1/2-(sqrt3)i]=7 By zeros we know there are roots around +sqrt7 and -sqrt7 and +1 (someone else found +-1.16) and -1 which can be used to start your iteration. I went through this exercise to show how many roots there are in this equation. Good job on your work.

    • 2 years ago
  38. benzilla Group Title
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    Oh sorry +-1/7 instead of +-1. See this makes the eyes blur. The best iterative program is to subtract the error if your over and determine a ratio multiplier that wont overshoot the answer. You can do this with a simple loop.

    • 2 years ago
  39. benzilla Group Title
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    I meant over and under. It will converge to the answer.

    • 2 years ago
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