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..

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).
