# Finding roots of cubic equation

If $\alpha,\beta,\gamma$ are the roots of the equation $2x^3-3x^2-12x+1=0$.Then find the value of [$\alpha$]+[$\beta$]+[$\gamma$],where [.] denotes greatest integer function.

My attempt:
I first tried hit and trial to guess the first root but no luck.Then I tried rational root method, but its roots are not rational (both roots could not satisfy the equation.)

I think its roots are irrational. Can someone help me finding its roots?

• – lab bhattacharjee Jul 26 '15 at 16:09
• – Macavity Jul 26 '15 at 16:12

Localise the roots between two consecutive integers and use the Intermediate value theorem.

Let $p(x)=2x^3-3x^2-12x+1$. As $p'(x)=6x^2-6x-12=6(x+1)(x-2)$, $p(x)$ has a local maximum at $x=-1$ and a local minimum at $x=2$: $$p(-1)=8,\quad p(2)=-19.$$ Thus we know there are three real roots: $x1<-1$, $-1< x_2 <2$, $x_3>2$. Now a table of the values of $p(x)$ at some integer points will do: $$\begin{matrix}x&p(x)\\\hline-2&-3\\-1&8\\0&1\\1&-12\\2&-19\\3&-8\\4&33\end{matrix}$$ Hence $\lfloor x_1\rfloor=-2$, $\lfloor x_2\rfloor=0$, $\lfloor x_3\rfloor=3$, so that $\lfloor x_1\rfloor+\lfloor x_2\rfloor+\lfloor x_3\rfloor=1$.

Let $$f(x)=2x^3-3x^2-12x+1$$

Now,

$f(0)>0$ and $f(1)<0 \implies 0< \alpha <1$

$f(-2)<0$ and $f(-1)>0 \implies -2< \beta <-1$

$f(3)<0$ and $f(4)>0 \implies 3< \gamma <4$

$\therefore \lfloor\alpha\rfloor + \lfloor\beta\rfloor + \lfloor\gamma\rfloor = 0+(-2)+3=1$

• This is clean, but it may be illustrative to clarify how you got -2,-1,0,1,3,4--you certainly didn't pull them from thin air, and doing so should deepen OP's understanding of how to approach such problems. – MichaelChirico Jul 26 '15 at 18:22
• The first strategy to try on problems like this is: look at easy small numbers. It is trivial to check $-1,0,1$. You immediately get a root between $0$ and $1$. Since $f(-1)$ is positive like $f(0)$, try $f(-2)$. That works. Since $f(1)<0$ the final root must be above 1. Try a few steps up to see Ishu has two simple typos in the next to last line, which do not prevent getting the answer. – Colin McLarty Jul 27 '15 at 6:40

Given $f(x)=2x^3-3x^2-12x+1$, we have a sign change between $-2$ and $-1$, between $0$ and $1$, between $3$ and $4$.