Question

Let $f\left(x\right)=x^2+6x+c,c\in R.$ If $f\left(f\right(x\left)\right)=0$ has exactly three distinct real roots, then the value of $c$ can be

• A. $9$
• B. $-3$
• C. $\frac{11-\sqrt{13}}{2}$
• D. $\frac{11+\sqrt{13}}{2}$

Answer 1

The correct option is C $\frac{11-\sqrt{13}}{2}$

$f\left(x\right)=x^2+6x+c,c\in R$
Since, $f\left(x\right)$ is a polynomial of degree $2,$
hence, $f\left(f\right(x\left)\right)$ will be a polynomial of degree $4.$

We know that, complex roots always occur in conjugate pair when co-efficients are real.
Since, $f\left(f\right(x\left)\right)=0$ has three distinct real roots, so one root must be repeated.

Let roots of $f\left(x\right)=0$ be $x=\alpha ,\beta$
So, $f\left(f\right(x\left)\right)=0$ has exactly three distinct real roots when $\alpha =\frac{-D}{4a}$ and $\beta >\frac{-D}{4a}$
$\Rightarrow \alpha =\frac{-(36-4c)}{4}=c-9$

Now, $f(-\frac{D}{4a})=0$
$\Rightarrow f(c-9)=0$
$\Rightarrow (c-9{)}^2+6(c-9)+c=0$
$\Rightarrow c^2-11c+27=0$
$\Rightarrow c=\frac{11\pm \sqrt{13}}{2}$

For $f\left(x\right)=x^2+6x+c=0,$ we have
$\alpha +\beta =-6$
$\Rightarrow \beta =3-c(\because \alpha =c-9)$

But, $\beta >-\frac{D}{4a}$
$\Rightarrow 3-c>c-9$
$\Rightarrow 2c<12$
$\Rightarrow c<6$
$\therefore c=\frac{11-\sqrt{13}}{2}$