Question

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. ?

Answer 1

`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`