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

Mathematics
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Not sure :p
|dw:1350910904638:dw|
That's also the point I reached. and then \(\rm \large x^2(x^{12 }-1) - 7 = 0\)

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Other answers:

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

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