= Complex conjugate roots in Vieta's formulas
For the <quadratic polynomial> $p(x)=2x^2-6x+5$, the <discriminant> is $(-6)^2-4(2)(5)=-4$, so its <roots of a polynomial> are the <complex conjugates>
$$
r_+=\frac{3+i}{2},\qquad r_-=\frac{3-i}{2}.
$$
Their <sum> is $3=-(-6)/2$, and their <product> is $(9+1)/4=5/2$, matching the constant <coefficient> divided by the <leading coefficient>. More revealingly, $r_+r_-=r_+\overline{r_+}=|r_+|^2$: the <product> is the square of a <complex modulus>. <Coefficients> in the <real numbers> therefore need not imply <polynomial roots> in the <real numbers>.
\Image[/vieta-complex-conjugate-roots.png]
{title=Complex conjugate roots of $2x^2-6x+5$ satisfy the same coefficient identities}
{description=The two stipulated <roots of a polynomial> are reflections across the <real axis> of the <complex plane>. Their <sum> cancels the <imaginary parts>, and their <product> is the squared <complex modulus> $5/2$. The pale arc marks that common <complex modulus>. Both positions belong in the root list even though there are no <roots of a polynomial> in the <real numbers>.}
For any <polynomial> with <coefficients> in the <real numbers>, $p(\overline z)=\overline{p(z)}$. Thus a nonreal <root of a polynomial> brings its <complex conjugate> along with the same <multiplicity of a root>. This follows by applying <complex conjugation> to the <polynomial> expression and its factors. It explains why sums of <complex conjugate> terms can produce <coefficients> in the <real numbers>.
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