Number of integers for which $3x^3-25x+n=0$ has three real roots

Question

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

Answer 1

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 $n$.

Answer 2

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.

Answer 3

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 $$|n| \leq \sqrt{50^3 \over 9^2 }= {250\sqrt{2}\over 9} <40$$