Question

Let $f\left(x\right)=x^2+5x+6$, then the number of real roots of $\left(f\right(x){)}^2+5f(x)+6-x=0$ is

• A. $2$
• B. $1$
• C. $3$
• D. $0$

Answer 1

The correct option is D

$0$

Let $f\left(x\right)=x$
$\Rightarrow x^2+5x+6=x$
$\Rightarrow x^2+4x+6=0$
Discriminant of the above equation:
$\Delta =4^2-4\left(1\right)\left(6\right)=-8$
Since the disciminant is negative, the equation $f\left(x\right)=x$ has non-real roots.

This means that the $f\left(f\right(x\left)\right)-x=0$ also has non-real roots.
$\Rightarrow \left(f\right(x){)}^2+5f(x)+6-x=0$ has $0$ real roots.