There is an alternate approach. If e + ig is a root and r + is is the other root, then a(x - e - ig)(x - r -is)=ax^2 + bx + c.
This then leads to
x^2 ] a=a
x ] a(-e -r - i(g + s)) = b, since b is real, s = -g
constant ] a(rc -gs -i(rg + es)) = c, since c is real, rg +es = 0 and since s = -g, r=e.
That is the same approach; that's how one derives the quadratic formula.
There's also a problem with this approach, what if
? Then the roots are both real, and the other root is NOT the complex conjugate, so again you need to assume the discriminant is negative to get your version. In this case you conclude the desired result by noting that g=0, so trivially e-gi is a root. It is vital that the question is phrased as "[. . .] e-gi is also a root." and NOT as "[. . .]e-gi is the OTHER root", since that is false.