SOLUTION: How do you solve for x for the equation 64x^4=x

Question 799145: How do you solve for x for the equation
64x^4=x
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
This is tricky, but that's why I chose to answer it. Please read the answer carefully so you become adept at solving all equations!
Given:
(1)
Don't be afraid of this "unusual" form. Just move the x to the left side to get
(2)
Look a little better now?
Now what's the first step in all "find x" problems?
Right, we FACTOR. So factor (2) into
(3)
Look better?
The second step is to set each factor equal to zero and solve for the root(s).
Set the first factor to zero yields
(4) x = 0, voila we have our first "solution".
Now set the second factor equal to zero and get
(5) or
(6) or
(7) or
(8) x = cube root of (1/4^3)
Since this is a cubic equation, there are three solutions, one real and two complex.
The real root of (8) is simply
(9) x = 1/4
The other two are complex. Are you familiar with complex numbers? I hope so.
If not, you should not be doing this advanced problem.
The three roots are equally spaced around a circle whose radius is the absolute value (1/4 in this case). A circle contains 360 degrees. Divided into three parts gives us an angle, say a, of
(10) a = 360/3 or
(11) a = 120 degrees.
Our first root of (8) is given by (9) which is real or at an angle of zero degrees, i.e. the real line labelled x on our graph.
The two complex roots can be expressed in polar form or complex form. In polar form we give the magnitude of the root and the angle at which it occurs (shown on the graph). In our case we have
(12) x = 1/4 at an angle of +120 degrees and
(13) x = 1/4 at an angle of -120 degrees.
The four solutions to the given fourth order equation (1) is at
(14) x = {0, 1/4, 1/4 @ +120 deg, 1/4 @ -120 deg} or in complex form
(15) x = {0, 1/4, (-1/8+sqrt(3)/8), (-1/8-sqrt(3)/8}

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