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sauravshakya

  • 2 years ago

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

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  1. ParthKohli
    • 2 years ago
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    Not sure :p

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

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

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

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

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

  7. benzilla
    • 2 years ago
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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

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

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

  10. UnkleRhaukus
    • 2 years ago
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    now what,

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

  12. UnkleRhaukus
    • 2 years ago
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    there are fourteen solutions

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

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

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

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

  17. sauravshakya
    • 2 years ago
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    Yep

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

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

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

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

  22. mukushla
    • 2 years ago
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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 :)

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

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

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

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

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

  28. mukushla
    • 2 years ago
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    trial and error

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

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

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

  32. benzilla
    • 2 years ago
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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.

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

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

  35. UnkleRhaukus
    • 2 years ago
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    hmm, good thinking

  36. benzilla
    • 2 years ago
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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

  37. benzilla
    • 2 years ago
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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.

  38. benzilla
    • 2 years ago
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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.

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

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