How do you find all the real and complex roots of #8-4x^3+4x^6=0#?

1 Answer
Apr 20, 2016

Answer:

Roots are three complex conjugate pairs #x =1.123 e^(+- 0.385pi), 1.123 e^(+- i 1.052 pi) and 1.123 e^ (+- i 1.719 pi)#, using 4-sd rounded approximations for irrational values..

Explanation:

The equation is a quadratic in #x^3#. Solving this quadratic
#4 (x^3)^2 - 4 x^3 + 8 = 0, x^3 = (1+- i sqrt7 )/2#

Now #x = ( (1+- i sqrt7 )/2)^(1/3) = (sqrt 2 e^((2kp+-1.209pi)i))^(1/3)

= 2^(1/6) e^((2kp+-1.209pi)i/3), k=0,1,2

= 1.123 e^(+- 0.385pi), 1.123 e^(+- i 1.052 pi) and 1.123 e^ (+- i 1.719 pi)#,

I have used #(1+isqrt7)/2=r(cos a + i sin a)

= r e^(ia) = sqrt2 e^(i 1.209pi)#. for generating all the roots..

Here, # a = cos^(-1)(1/sqrt.8)

=69.30^o=69.30/180 pi = 0.385pi# radian