At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

Not sure :p

|dw:1350910904638:dw|

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

And then I factored it by difference of squares continuously.

Then I reached nothing.

|dw:1350910926540:dw|

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

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.

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

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?

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

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