How do you solve x^2 + 2 = 0 and x^4 + 4 = 0?

[Meh]

Find all real/imaginary roots to

x^9=16x

 

[mathmike]

have you tried to solve this yourself?

try completing the square

 

[mathmike]

or maybe a diff of squares

 

[Meh]

I'm stuck right after i bring the 16x over...
x^9-16x=0

Where do I go from here?I have tried this, it was a test question for me today. Didn't get it so just wondering what the answer is.

 

[Tide]

HINT 1: Factor! :)

HINT 2: Think Euler!

 

[Meh]

Can someone just give the answer? I don't got a clue on how to factor it : (

 

[quantumdude]

No, we will not just give the answer.

Start from your equation: x^9-16x=0

What's the first thing you should look for when factoring? A common factor.

 

[mathelord]

x[x^8 -16]=0
x=0,x^8=16 now solve the latter

 

[Diane_]

mathelord - If he does it your way, he'll miss some roots - all of the complex ones, actually.

Meh - do as Tide suggested. Factoring is the way to go. Let me suggest you go back and review some of the basic factoring patterns - sum of cubes, difference of cubes, things of that nature.

 

[ivybond]

Meh - if you've dealt with polynomial equations before, you might remember that a polynom of n-th degree has n roots (real or complex or combination of both).
So P₈(x) = x⁸ - 16 has 8 roots.

One way of finding them is applying a very useful DeMoivre's Theorem to
x⁸ = 16
and extracting a root of 8th degree.
If it's not in your course, it's really worth mastering.
If you do, it will give you a serious sense of satisfaction.

Otherwise it can be done the way Tide and Diane_ suggested, except that Tide's "Euler hint" may not be needed.

 

[borisleprof]

Do you know how to solve
x^2 + 2 = 0 ?
and
x^4 + 4 = 0?
Look at the first one.
(x+i√2)(x-i√2)=0
What for the second x^4 + 4=x^4+4x^2+4-4x^2;
maybe you can continue and find your answer