anonymous
  • anonymous
Another Fun Question Find all the real solutions: x^14=7+x^2
Mathematics
schrodinger
  • schrodinger
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ParthKohli
  • ParthKohli
Not sure :p
ksaimouli
  • ksaimouli
|dw:1350910904638:dw|
ParthKohli
  • ParthKohli
That's also the point I reached. and then \(\rm \large x^2(x^{12 }-1) - 7 = 0\)

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ParthKohli
  • ParthKohli
And then I factored it by difference of squares continuously.
ParthKohli
  • ParthKohli
Then I reached nothing.
ksaimouli
  • ksaimouli
|dw:1350910926540:dw|
anonymous
  • anonymous
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
UnkleRhaukus
  • UnkleRhaukus
\[x^{14}=7+x^2\]\[x^{14}-x^2=7\]\[x^2(x^{12}-1)=7\]
anonymous
  • anonymous
oops i made a mistake I forgot the 7
UnkleRhaukus
  • UnkleRhaukus
now what,
anonymous
  • anonymous
The equation is then xX2(x-1)(x^11+x^10,,,,,,)=7
UnkleRhaukus
  • UnkleRhaukus
there are fourteen solutions
anonymous
  • anonymous
Just find real solutions.
anonymous
  • anonymous
I set this to zero which made this come out in error. This needs an iterative process to solve.
anonymous
  • anonymous
Yep... @benzilla
anonymous
  • anonymous
if there are some answers they are in the range\[\rm 1<|x|<2\]
anonymous
  • anonymous
Yep
anonymous
  • anonymous
answer will be close to 1 so letting\[x^2 \approx1\]\[x^{14}=8\]\[x=\sqrt[14]{8}\]
anonymous
  • anonymous
actually\[x\approx\pm\sqrt[14]{8}\]
anonymous
  • anonymous
What kind of solution is that???
anonymous
  • anonymous
thats just a approximate solution...not exact
anonymous
  • anonymous
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 :)
anonymous
  • anonymous
Yep... What about the exact solution.
anonymous
  • anonymous
I think it is irrational
anonymous
  • anonymous
well thats what engineers do...lets think like mathematicians
anonymous
  • anonymous
Let y=x^2 then, y^7=7+y y=(7+y)^(1/7)
anonymous
  • anonymous
|dw:1350912581733:dw|
anonymous
  • anonymous
trial and error
anonymous
  • anonymous
And that gives:|dw:1350912666885:dw|
anonymous
  • anonymous
Thus, |dw:1350912725583:dw|
anonymous
  • anonymous
Can I do that?
anonymous
  • anonymous
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.
ganeshie8
  • ganeshie8
thats very much an exact solution :)
anonymous
  • anonymous
And @UnkleRhaukus from similar process the solution of x^3=1+x is |dw:1350913011409:dw|
UnkleRhaukus
  • UnkleRhaukus
hmm, good thinking
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
I meant over and under. It will converge to the answer.

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