Thread: Solving Quartic

Find all roots of x^4 + i =0. Write answers in trigonometric form.

Don't know how to solve quartics.
Any help would be appreciated.

Originally Posted by c_323_h

Find all roots of x^4 + i =0. Write answers in trigonometric form.

Don't know how to solve quartics.
Any help would be appreciated.

You need to use a theorem know as De' Moivre's
.
You need to express

as,

Now, express in trigonomteric form,

Thus,

Thus,

By De Moivre's Theorem,

But, this is only one solution there are 3 more (because of the degree). In this case, we divide, by the degree, (4) to get, and each other solution is added by
Thus,




If you want you can add to your first solution to get,






The beauty of the second result is that it is very easy to see, that each one is divided by 8 and the numerator is increasing by 4.

Thanks!

when you solve for roots do you always solve for x?

Originally Posted by c_323_h

Thanks!

when you solve for roots do you always solve for x?

Not sure I understand your question

You subracted from both sides to get

Why did you subtract ?

Originally Posted by c_323_h

You subracted from both sides to derive

Why did you subtract ?

You mean, in a regular poynomial you factor one side so you can set its factors equal to zero but over here you do not do this?

Because a special equation of polynomial is,
.
We can show that they are EXACTLY solutions and there are called Roots of Unity which are of quite importance in Abstract Algebra/Field Theory.

Whenever, you see and equation like this you bring the constant term to the other side and then use De Moivre's Theorem to find its roots.
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It is interesting to note, that the soltuions that you get, are in trigonomteric form. Some of them can be simplified in algebraic terms (only using +,-,x,/ and roots) while some of them cannot (impossible) and thus you need to leave it in the form it is in.