## sauravshakya Group Title Another Fun Question Find all the real solutions: x^14=7+x^2 one year ago one year ago

1. ParthKohli Group Title

Not sure :p

2. ksaimouli Group Title

|dw:1350910904638:dw|

3. ParthKohli Group Title

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

4. ParthKohli Group Title

And then I factored it by difference of squares continuously.

5. ParthKohli Group Title

Then I reached nothing.

6. ksaimouli Group Title

|dw:1350910926540:dw|

7. benzilla Group Title

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 Group Title

$x^{14}=7+x^2$$x^{14}-x^2=7$$x^2(x^{12}-1)=7$

9. benzilla Group Title

oops i made a mistake I forgot the 7

10. UnkleRhaukus Group Title

now what,

11. benzilla Group Title

The equation is then xX2(x-1)(x^11+x^10,,,,,,)=7

12. UnkleRhaukus Group Title

there are fourteen solutions

13. sauravshakya Group Title

Just find real solutions.

14. benzilla Group Title

I set this to zero which made this come out in error. This needs an iterative process to solve.

15. sauravshakya Group Title

Yep... @benzilla

16. mukushla Group Title

if there are some answers they are in the range$\rm 1<|x|<2$

17. sauravshakya Group Title

Yep

18. mukushla Group Title

answer will be close to 1 so letting$x^2 \approx1$$x^{14}=8$$x=\sqrt[14]{8}$

19. mukushla Group Title

actually$x\approx\pm\sqrt[14]{8}$

20. sauravshakya Group Title

What kind of solution is that???

21. mukushla Group Title

thats just a approximate solution...not exact

22. mukushla Group Title

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 Group Title

Yep... What about the exact solution.

24. sauravshakya Group Title

I think it is irrational

25. mukushla Group Title

well thats what engineers do...lets think like mathematicians

26. sauravshakya Group Title

Let y=x^2 then, y^7=7+y y=(7+y)^(1/7)

27. sauravshakya Group Title

|dw:1350912581733:dw|

28. mukushla Group Title

trial and error

29. sauravshakya Group Title

And that gives:|dw:1350912666885:dw|

30. sauravshakya Group Title

Thus, |dw:1350912725583:dw|

31. sauravshakya Group Title

Can I do that?

32. benzilla Group Title

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 Group Title

thats very much an exact solution :)

34. sauravshakya Group Title

And @UnkleRhaukus from similar process the solution of x^3=1+x is |dw:1350913011409:dw|

35. UnkleRhaukus Group Title

hmm, good thinking

36. benzilla Group Title

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 Group Title

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 Group Title

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 Group Title

I meant over and under. It will converge to the answer.