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..
Best Answer
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 $\pm 2$.