Number of integers for which $3x^3-25x+n=0$ has three real roots Number of integers n for which $3x^3-25x+n=0$ has three real roots then
a)$1$
b)$25$
c)$55$
d)infinite
This appeared in an national level examination .
Now this may be solved graph.
I have actually really no idea how to this and according to me the answer should be infinite
 A: Let cubic $f(x) = 3x^3 -25x$ and horizontal line $g(x) =-n$. 
The approach here would be to find derivative $f'(x)$ and check it's roots. 
Here $f'(x)$ has two real and distinct roots, $a,b$. This means that $f(x)=g(x)$ will have three real roots if the horizontal line  $g(x) \in [f(a), f(b)]$
The roots of $f'(x)$ are $\pm\frac{5}{3}$. So we have $-n \in [f(\frac{-5}{3}), f(\frac{5}{3})]$.
$$-n\in\left[\tfrac{-250}{9}, \tfrac{250}{9}\right]$$
This gives $55$ solutions for $n$.
A: More generally, a cubic equation $f(x)=c$ has three real roots for all values of $c$ between $f(x_1)$ and $f(x_2)$, where $x_1$ and $x_2$ are the critical points of $f$.
For $f(x)=3x^3-25x$, the critical points are $\pm 5/3$ and the interval is $(-250/9, 250/9) \approx (-27.7, 27.7)$, which contains $55$ integers.
Alternatively, you can use this fact

A cubic equation has three distinct real roots iff its discriminant is positive.

The discriminant of $3x^3-25x+n$ is $\Delta=-3 (81 n^2 - 62500)$. Therefore, $\Delta>0$ iff $81 n^2 - 62500<0$, that is, iff $n \in (-250/9, 250/9)$, as before.
A: By Vieta's formulas we have:
$$ x_1+x_2+x_3=0$$
$$ x_1x_2+x_2x_3+x_3x_1 =-25/3$$
So $$x_1^2+x_2^2+x_3^2=50/3$$ and because $$3\sqrt[3]{x_1^2x_2^2x_3^2} \leq x_1^2+x_2^2+x_3^2$$
and since $x_1x_2x_3 = -n/3$ we have $$3\sqrt[3]{n^2/9} \leq 50/3$$
So $$|n| \leq \sqrt{50^3 \over 9^2 }= {250\sqrt{2}\over 9} <40$$
