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?
$\endgroup$ 3Tried 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.
4 Answers
$\begingroup$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 $m+n+r=200$
$\endgroup$ 2 $\begingroup$HINT: try the ansatz $$(-2+Bx+Ax^2)(1+Cx^2+Dx^4)$$
$\endgroup$ 9 $\begingroup$$$2000x^6+100x^5-200x^4+200x^4+10x^3-20x^2+20x^2+x-2$$ $$(2000x^6+200x^4+20x^2)+(100x^5+10x^3+x)-(200x^4+20x^2+2)$$ $$x^2(2000x^4+200x^2+20)+\frac{x}{20}(2000x^4+200x^3+20x)-\frac{1}{10}(2000x^4+200x^2+20)$$
$$=(x^2+\frac{x}{20}-\frac{1}{10})(2000x^4+200x^2+20)=0$$
$\endgroup$ $\begingroup$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 $$ 20 x^2 + x - 2 = 0 $$
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