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.

• All roots or only real roots ? – Claude Leibovici Jun 20 '15 at 12:43
• Is this problem supposed to be solved by hand? – ajotatxe Jun 20 '15 at 12:48
• Wolphram Alpha gives the factors $(20x^2+x-2)(100x^4+10x^2+1)$ so you can finish easily with $x^2=Z$ – Piquito Jun 20 '15 at 13:40

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 $m+n+r=200$

• Very nice, +1! One minor nit: it might be worth showing why the $1000x^6-1=0$ path doesn't work out (it leads to a couple of complex solutions plus $x^2=\frac1{10}$, and even without plugging in to test whether these solve the original sextic it's clear that it can't be put into the required form.) – Steven Stadnicki Jan 28 '14 at 19:36
• This equation will not give the roots in the required form.And if you are thinking of complex solutions in this form check out the conditions for m,n,r(they should be positive integers) – Devgeet Patel Jan 29 '14 at 3:15

HINT: try the ansatz $$(-2+Bx+Ax^2)(1+Cx^2+Dx^4)$$

• and thankyou for downvoting – Dr. Sonnhard Graubner Jun 20 '15 at 12:50
• and this here $$(20x^2+x-2)(100x^4+10x^2+1)$$?? – Dr. Sonnhard Graubner Jun 20 '15 at 12:58
• Such a system of nonlinear equations does not seem easy to manipulate. Could you elaborate ? – Claude Leibovici Jun 20 '15 at 12:58
• this is the answer a method which is always used in math olympiads – Dr. Sonnhard Graubner Jun 20 '15 at 13:04
• I deleted a bunch of repetitive comments. Gentlemen, remember that this is a sextic. Either there is a trick to find the zeros, or they cannot be found. So we should expect an ad hoc -method. – Jyrki Lahtonen Jun 20 '15 at 13:25

$$2000x^6+100x^5-200x^4+200x^4+10x^3-20x^2+20x^2+x-2$$ $$(2000x^6+200x^4+20x^2)+(100x^5+10x^3+x)-(200x^4+20x^2+2)$$ $$x^2(2000x^4+200x^2+20)+\frac{x}{20}(2000x^4+200x^3+20x)-\frac{1}{10}(2000x^4+200x^2+20)$$

$$=(x^2+\frac{x}{20}-\frac{1}{10})(2000x^4+200x^2+20)=0$$

The usual trick is to divide through by $x^3.$ Then notice that taking $w = 10x - \frac{1}{x}$ seems to allow writing the thing, and we get $$2 w^3 + w^2 + 60 w + 30 = 0.$$ This has a rational root, namely $-1/2,$ and factors as $$(2w+1)(w^2 + 30)$$ Setting $$10 x - \frac{1}{x} = -\frac{1}{2}$$ leads to $$20 x^2 + x - 2 = 0$$