Question

how to solve this, complex roots
(x / 2) ^ 6 - 1 = 0

Answer 1

Hey Anna! :)
Welcome to OpenStudy.

So you have a couple of options here.
You can simply factor the difference of cubes down,
Which might be a little tedious.

Or you can convert 1 to polar and use De'Moivre's Theorem or something.

Here is the difference of cubes formula in case you forgot:\[\Large\rm a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 2

\[\Large\rm \left[\left(\frac{x}{2}\right)^2\right]^3-1^3=0\]So we have the difference of cubes, yes?

Answer 3

Comeon Anna! Wake up!
Lemme know what you're thinking! c:

Answer 4

other example is :
x^5+16-0 so we have x*(x^4+16)=0
x=0 and x^4+16=0 so x^4=-16
p=(cos+sin)=-16
etc.. after that im finding roots
my problrm is with the form (x/2)^6 - 1 = 0 how to simplify this form??

Answer 5

So I started by applying an exponent rule which allowed me to break up the 6 into 2 and 3.
Next we apply the difference of cubes formula which gives us:\[\Large\rm \left[\left(\frac{x}{2}\right)^2-1\right]\left[\left(\frac{x}{2}\right)^4+\left(\frac{x}{2}\right)^2+1\right]=0\]I guess that's the really tricky step in all of this^
being able to apply the difference of cubes correctly.


So the first factor, since it's subtraction, will give us our real solutions.
\[\Large\rm \left(\frac{x}{2}\right)^2=1\]Square root each side, multiply by 2,\[\Large\rm x=\pm 2\]
For the other factor, make a substitution maybe...
\[\Large\rm u=\left(\frac{x}{2}\right)^2\]
Which leads us to,\[\Large\rm u^2+u+1=0\]

Answer 6

And then from there... umm... thinking..

Answer 7

Applying your Quadratic Formula will get you to:\[\Large\rm u=\frac{-1\pm\sqrt{1-4}}{2}\]\[\Large\rm u=-\frac{1}{2}\pm \frac{\sqrt3}{2}\mathcal i\]Substituting our u back out,\[\Large\rm \left(\frac{x}{2}\right)^2=-\frac{1}{2}\pm \frac{\sqrt3}{2}\mathcal i\]Square rooting each side, multiplying by 2 gives us,\[\Large\rm x=\pm2\left(-\frac{1}{2}\pm\frac{\sqrt3}{2}\mathcal i\right)^{1/2}\]

Answer 8

And then ummm....

Answer 9

I'm not quite sure what you were doing with your `p` thing.
But I guess you could write this as two separate function for the pm on sqrt3/2,\[\Large\rm x=\pm2\left(-\frac{1}{2}+\frac{\sqrt3}{2}\mathcal i\right)^{1/2}\]And,\[\Large\rm x=\pm2\left(-\frac{1}{2}-\frac{\sqrt3}{2}\mathcal i\right)^{1/2}\]And then you can convert to sines and cosines from there.
That doesn't seem to match what your example did though...

I don't really like the approach I took.
Complex exponentials is way way easier.
I'm not exactly sure where your understanding of this is, so it's hard to pinpoint how we should do this.