Question

Let $f\left(x\right)=a{x}^2+bx+c$ where $a,b,c\in R$. If $f\left(x\right)$ takes real values for real values of $x$ and non-real values for non-real values of $x$, then

• A. $a=0$
• B. $b=0$
• C. $c=0$
• D. nothing can be said about $a,b,c$

Answer 1

The correct option is A $a=0$

It might so happen that for a non real $x$, $f\left(x\right)$ may be real for some values of $a$ and $b$.

Suppose $a\ne 0.$ We rewrite $f\left(x\right)$ as follows:

$f\left(x\right)=a\{x^2+\frac{b}{a}x+\frac{c}{a}\}$

$=a\{{(x+\frac{b}{2a})}^2+\frac{4ac-b^2}{4a^2}\}$

$f(-\frac{b}{2a}+i)=a\{{(-\frac{b}{2a}+i+\frac{b}{2a})}^2+\frac{4ac-b^2}{4a^2}\}$

$=a\{-1+\frac{4ac-b^2}{4a^2}\},$ which is a real number

This is against the hypothesis. Therefore, $a=0$.