# Find the roots of cubic equation.

If the function $$f(x)=x^3-9x^2+24x+c$$ has three real and distinct roots $l,m,n$, the find value of $[l]+[m]+[n]$ where $[..]$ represents greatest integer function

In my book there is no information given about $c$.

• Do you know the condidtion for real roots – Archis Welankar May 5 '16 at 11:41
• It has stationary points at $x=2,4$ (differentiate and factorise), so you need the value at $x=2$ to be positive and the value at $x=4$ to be negative. That gives $-16<c<-20$. You should find $[l]+[m]+[n]$ is 7 or 8. – almagest May 5 '16 at 11:49
• @almagest can you please tell why value has to be positive at $x=2$ and negative at $x=4$ – ramsay May 5 '16 at 15:52
• The local maximum is at $x=2$ and the local minimum at $x=4$. If the $x$-axis comes above the local maximum, then there is only one real root. Similarly, if it is below the local minimum, there is only one real root. For three real roots you need it to be between the two. Hence you need $f(2)>0$ and $f(4)<0$. Drawing a rough graph might help. – almagest May 5 '16 at 15:57
• @almagest my bad! but say you say why it is local maximum at 2 and l. minimum at 4. unless you don't know 'c', how can you say that? – ramsay May 5 '16 at 16:02

HINT

First you can establish that the graph must have stationary points at $x=2$ and $x=4$

For there to be three real distinct roots, you require the values of $f(2)$ and $f(4)$ to be of opposite parity.

This gives rise to the condition that $-20<c<-16$

For each of the extreme values of $c$ the polynomial factorises nicely and from a sketch you will be able to deduce the quantity you are looking for.

I hope this helps

• "$f(2)$ and $f(4)$ to be of opposite parity.", but why? can't they be humped upwards in both $f(2)$ and $f(4)$ – ramsay May 5 '16 at 15:51

The discriminant of this cubic is $-27 c^2 - 972 c - 8640 = -27(c+20)(c+16)$. To have three distinct real roots, we need the discriminant positive, so $-20 < c < -16$.

Note that the values of $f$ at $x = 1,2,3,4,5$ are $c+16, c+20, c+18,c+16,c+20$ respectively.

For $c$ between $-20$ and $-16$, the lowest root is between $1$ and $2$, the highest between $4$ and $5$. The middle root is between $2$ and $4$, is an increasing function of $c$, and is $3$ when $c=-18$. So for $-20 < c < -18$ we have $\lfloor l \rfloor + \lfloor m\rfloor + \lfloor n\rfloor = 1+2+4=7$, and for $-18 \le c < -16$ we have $1+3+4 = 8$.

From Vieta's formula:

$$l+m+n=9$$

If $l,m,n\in\mathbb R$, then

$$\lfloor l\rfloor+\lfloor m\rfloor+\lfloor n\rfloor\ge l+m+n=9$$

Therefore, we have

$$6\le\lfloor l\rfloor+\lfloor m\rfloor+\lfloor n\rfloor\le9$$

Since we have local maximum and minimum at $x=2,4$, then

$$f(2)=20+c$$

$$f(4)=16+c$$

Therefore, $f(2)>f(4)$.

• The OP wants $\lfloor \ldots \rfloor$, not $\lceil \ldots \rceil$. – Robert Israel Jun 1 '16 at 22:10
• @RobertIsrael Oh, my bad. – Simply Beautiful Art Jun 1 '16 at 22:11