If #1+3^(1/2) i# is the root of the equation #2z^3+az^2+bz+4=0#,find the values of the real numbers of a and b. For the values of a and b, solve the equation. ?

1 Answer
George C.
Feb 27, 2018

#a=-3#, #b=6# and the roots are:

#1+sqrt(3)i#, #1-sqrt(3)i# and #-1/2#

Explanation:

Given that #1+sqrt(3)i# is a root of:

#2z^3+az^2+bz+4 = 0#

and the coefficients are real, #1-sqrt(3)i# must also be a root and we have a factor:

#(z-1-sqrt(3)i)(z-1+sqrt(3)i) = (z-1)^2-(sqrt(3)i)^2#

#color(white)((z-1-sqrt(3)i)(z-1+sqrt(3)i)) = z^2-2z+1+3#

#color(white)((z-1-sqrt(3)i)(z-1+sqrt(3)i)) = z^2-2z+4#

The remaining linear factor must have leading term #2z# and trailing term #1# in order that the leading and trailing terms of the product are #2z^3# and #4#.

So:

#0 = (2z+1)(z^2-2z+4) = 2z^3-3z^2+6z+4#

So #a=-3#, #b=6# and the third root is #z=-1/2#

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