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ksaimouliBest ResponseYou've already chosen the best response.2
dw:1350910904638:dw
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
That's also the point I reached. and then \(\rm \large x^2(x^{12 }1)  7 = 0\)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
And then I factored it by difference of squares continuously.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Then I reached nothing.
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.2
dw:1350910926540:dw
 one year ago

benzillaBest ResponseYou've already chosen the best response.0
X^14 x^2=7..........>X^2(X^121)...........>x^2(x1)(x^11+x^10.......) The second expression is prime so your two real roots are zero and one
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[x^{14}=7+x^2\]\[x^{14}x^2=7\]\[x^2(x^{12}1)=7\]
 one year ago

benzillaBest ResponseYou've already chosen the best response.0
oops i made a mistake I forgot the 7
 one year ago

benzillaBest ResponseYou've already chosen the best response.0
The equation is then xX2(x1)(x^11+x^10,,,,,,)=7
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
there are fourteen solutions
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.3
Just find real solutions.
 one year ago

benzillaBest ResponseYou've already chosen the best response.0
I set this to zero which made this come out in error. This needs an iterative process to solve.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
if there are some answers they are in the range\[\rm 1<x<2\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
answer will be close to 1 so letting\[x^2 \approx1\]\[x^{14}=8\]\[x=\sqrt[14]{8}\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
actually\[x\approx\pm\sqrt[14]{8}\]
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.3
What kind of solution is that???
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
thats just a approximate solution...not exact
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
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 :)
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.3
Yep... What about the exact solution.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.3
I think it is irrational
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
well thats what engineers do...lets think like mathematicians
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.3
Let y=x^2 then, y^7=7+y y=(7+y)^(1/7)
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.3
dw:1350912581733:dw
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.3
And that gives:dw:1350912666885:dw
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.3
Thus, dw:1350912725583:dw
 one year ago

benzillaBest ResponseYou've already chosen the best response.0
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.
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.1
thats very much an exact solution :)
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.3
And @UnkleRhaukus from similar process the solution of x^3=1+x is dw:1350913011409:dw
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
hmm, good thinking
 one year ago

benzillaBest ResponseYou've already chosen the best response.0
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^3i) (x1)[X1/2[(sqrt3)/2]i] [X1/2+[(sqrt3)/2]i](x+1) [X+1/2+[(sqrt3)/2]i] [X1/2[(sqrt3)/2]i]=7 X^2(X+i)(?)(Xi)(??) (x1)[X1/2[(sqrt3)/2]i] [X1/2+[(sqrt3)/2]i](x+1) [X+1/2+[(sqrt3)/2]i] [X1/2[(sqrt3)/2]i]=7 The ? and ?? can be factored however i am running out of time this morning. Good luck
 one year ago

benzillaBest ResponseYou've already chosen the best response.0
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)(Xi)(Xi(1+sqrt3)/2)(Xi(1sqrt3)/2) (x1)[X+1/2(sqrt3)i] [X+1/2+(sqrt3)i](x+1) [X1/2+(sqrt3)i] [X1/2(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.
 one year ago

benzillaBest ResponseYou've already chosen the best response.0
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.
 one year ago

benzillaBest ResponseYou've already chosen the best response.0
I meant over and under. It will converge to the answer.
 one year ago
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