# 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 $4$ distinct real roots, if f(k) =0, $k \neq \alpha^2, \beta^2$ ( where $\pm \alpha, \pm \beta)$ are roots of equation $f(x^2)=0,$ then ( which of the following is correct)

(a) $k <0$

(b) $k >0$

(c) k $\leq 0$

(d) $k \geq 0$

Please suggest how to get this relation of k w.r.t the given polynomial. I am having no clue how to proceed to get the options... will be greatful to you... thanks....

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Do you mean 'if $f(k)=0$'? –  Ian Coley Oct 23 '13 at 2:22
Hint: $k$ is a root of $f(x)$, but there is no corresponding root of $f(x^2)$. Therefore, $k$ cannot be written as the square of a real number. What does this tell you about $k$?
Is it $k \leq 0$ ? –  sultan Oct 23 '13 at 2:57
@sultan: No. If $f(0) = 0$, then $f(0^2) = 0$, so $0$ is a root of $f(x^2)$. Close, however. –  William Ballinger Oct 24 '13 at 3:55