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

- anonymous

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

- Stacey Warren - Expert brainly.com

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- schrodinger

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- ParthKohli

Not sure :p

- ksaimouli

|dw:1350910904638:dw|

- ParthKohli

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

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## More answers

- ParthKohli

And then I factored it by difference of squares continuously.

- ParthKohli

Then I reached nothing.

- ksaimouli

|dw:1350910926540:dw|

- 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

\[x^{14}=7+x^2\]\[x^{14}-x^2=7\]\[x^2(x^{12}-1)=7\]

- anonymous

oops i made a mistake
I forgot the 7

- UnkleRhaukus

now what,

- anonymous

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

- UnkleRhaukus

there are fourteen solutions

- anonymous

Just find real solutions.

- anonymous

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

- anonymous

Yep... @benzilla

- anonymous

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

- anonymous

Yep

- anonymous

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

- anonymous

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

- anonymous

What kind of solution is that???

- anonymous

thats just a approximate solution...not exact

- 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

Yep... What about the exact solution.

- anonymous

I think it is irrational

- anonymous

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

- anonymous

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

- anonymous

|dw:1350912581733:dw|

- anonymous

trial and error

- anonymous

And that gives:|dw:1350912666885:dw|

- anonymous

Thus, |dw:1350912725583:dw|

- anonymous

Can I do that?

- 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

thats very much an exact solution :)

- anonymous

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

- UnkleRhaukus

hmm, good thinking

- 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

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

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

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

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