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

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ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1350910904638:dw

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

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0And then I factored it by difference of squares continuously.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Then I reached nothing.

ksaimouli
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1350910926540:dw

benzilla
 3 years ago
Best ResponseYou've already chosen the best response.0X^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

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0\[x^{14}=7+x^2\]\[x^{14}x^2=7\]\[x^2(x^{12}1)=7\]

benzilla
 3 years ago
Best ResponseYou've already chosen the best response.0oops i made a mistake I forgot the 7

benzilla
 3 years ago
Best ResponseYou've already chosen the best response.0The equation is then xX2(x1)(x^11+x^10,,,,,,)=7

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0there are fourteen solutions

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.3Just find real solutions.

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

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0if there are some answers they are in the range\[\rm 1<x<2\]

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

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0actually\[x\approx\pm\sqrt[14]{8}\]

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.3What kind of solution is that???

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0thats just a approximate solution...not exact

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0we 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 :)

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.3Yep... What about the exact solution.

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.3I think it is irrational

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0well thats what engineers do...lets think like mathematicians

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

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.3dw:1350912581733:dw

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.3And that gives:dw:1350912666885:dw

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.3Thus, dw:1350912725583:dw

benzilla
 3 years ago
Best ResponseYou've already chosen the best response.0Beyond 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
 3 years ago
Best ResponseYou've already chosen the best response.1thats very much an exact solution :)

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

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0hmm, good thinking

benzilla
 3 years ago
Best ResponseYou've already chosen the best response.0If 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

benzilla
 3 years ago
Best ResponseYou've already chosen the best response.0Here 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.

benzilla
 3 years ago
Best ResponseYou've already chosen the best response.0Oh 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.

benzilla
 3 years ago
Best ResponseYou've already chosen the best response.0I meant over and under. It will converge to the answer.
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