If $a,b,c(a,b,c\in\mathbb{R} )$ satisfy $b^2-4ac<0$ then equation $f(x)=0$ has complex root

Question

I would appreciate if somebody could help me with the following problem:

Q: show that ($n>2, n\in\mathbb{N}$)

Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+ax^2+bx+c, ~(a_i\in\mathbb{R}(i=3,4,\cdots,n),a_n\neq 0) $

If $a,b,c(a,b,c\in\mathbb{R} )$ satisfy $b^2-4ac<0$ then equation $f(x)=0$ has complex root

Answer 1

It can't be helped because it is wrong. Indeed, just take the two examples of polynomials with all their roots being real $$ (x-1)(x-2)(x+3)=x^3-7x+6=0,\quad b^2-4ac=49>0, $$ $$ (x-1)(x-2)(x-3)=x^3-6x^2+11x-6=0,\quad b^2-4ac=-23<0, $$ despite the sign of the discriminant $b^2-4ac$.

Answer 2

If you mean that $f(x)=0$ must a have a complex root that isn't real, then this is not true. Look at the polynomial $$ f(x) = x^3 + 3x^2 + 3x + 1. $$

Here $b^2 - 4ac = 3^2 - 4\cdot 3 \cdot 1 = (-3) < 0$, but the only complex root of $f$ is $x_0 = -1$ (with multiplicity $3$), which is real.

If you mean that $f(x)=0$ must simply have a complex root, then this is true because $a_n \neq 0$ by the fundamental theorem of algebra.