Let $f(x)$ be a 3rd degree polynomial such that $f(x^2)=0$ has exactly $4$ distinct real roots Problem : 
Let $f(x)$ be a 3rd degree polynomial such that $f(x^2)=0$ has exactly four distinct real roots, then which of the following options are correct : 
(a) $f(x) =0$ has all three real roots 
(b) $f(x) =0$  has exactly two real roots 
(c) $f(x) =0$ has only one real root 
(d) none of these 
My approach : 
Since $f(x^2)=0 $ has exactly four distinct real roots. Therefore the remaining two roots left [ as $ f(x^2)=0$ is a degree six polynomial].
How can we say that the remaining two roots will be real . This will not have one real root ( as non real roots comes in conjugate pairs). 
So, option (c) is incorrect. I think the answer lies in option (a) or (b) . But I am not confirm which one is correct. Please suggest.. thanks..
 A: No requirement for $f$ to have real coefficients was stated.  So you could have
e.g. $f(x) = (x-1)(x-4)(x+i)$ which has two real roots, and $f(x^2)$ has the four distinct real roots $\pm 1$ and $\pm 2$.
A: I'm editing this because found a fatal error on the demonstration assuming $C\in\mathbb{R}$ witch may not be always true, leading to a wrong answer. Excuse me :/
First condition: $f(x)$ is 3rd degree. So, $f(x)$ can be expressed as $(x-A)(x-B)(x-C)$ with $A,B,C \in \mathbb{C}$. 
*Note that if $A \in \mathbb{C}$, it's possible that $A\in\mathbb{R}$.
If $f(x)=(x-A)(x-B)(x-C)$ then $f(x^2)=(x^2-A)(x^2-B)(x^2-C)$
Second condition: $f(x^2)$ has exactly four distinct real roots. It means that $A \ge 0, A\ne B\ge0, C \in (\mathbb{C}-\mathbb{R^+})$. I'm Assuming $0 \in \mathbb{R^+}$
Answer: d)
Cannot be a) because if $C \in \mathbb{C-R}$, $x=C$ is complex, so $f(x)$ does not have 3 real roots.
Cannot be b) because if $C \in \mathbb{R}$, $x=C$ is real, witch is the third root.
Cannot be c) because $A$ and $B$ are real roots of the equation.
Is d) because it says it cannot be a) b) and c).
A: Both $f(x)=(x-1)(x-4)^2$ (two real roots including one double root), and $f(x)=(x+1)(x-1)(x-4)$ (three real roots) fulfit the conditions of the problem, with $f(x^2)=0$ having roots in $x=\pm1$ and $x=\pm2$.
A: I assume that the polynomial has real coefficients.
Now if $f(x^2)$ has four distinct real roots, then the square of these roots are the roots of $f(x)$. Hence the polynomial has at least two distinct real roots. Hence it has three distinct real roots (as complex roots occur in pairs).
