Solve $2000x^6+100x^5+10x^3+x-2=0$ One of the roots of the equation $2000x^6+100x^5+10x^3+x-2=0$ is of the form $\frac{m+\sqrt{n}}r$, where $m$ is a non-zero integer and $n$ and $r$ are relatively prime integers.Then the value of $m+n+r$ is?

Tried to use the fact that another root will be $\frac{m-\sqrt{n}}r$ as coefficients are rational but there are six roots and using sum and product formulas would allow many variables in the equations.


We have $\displaystyle x+10x^3+100x^5=x\frac{1000x^6-1}{10x^2-1}$. (A geometric progression)
Hence $\displaystyle -2(1000x^6-1)=x \frac{1000x^6-1}{10x^2-1}$
Hence either $1000x^6-1=0$ or $x=-2(10x^2-1)$.
Therefore $20x^2+x-2=0$ for second equation.
Solving we get 
$$x=\frac{-1\pm \sqrt{161}}{40}$$.
Comparing $m=-1, n=161$ and $r=40$. Hence $m+n+r=200$
