Question

One of the roots of equation 2000x^6 +100x^5 +10x^3 +x -2 =0 is of form
\[\frac{m+\sqrt{n}}{r}\]
where m is non zero integer n and r are relatively prime natural nos.FInd value of m+n+r

Answer 1

Don't Cheat

Answer 2

Donot wolfram it

Answer 3

I won't :-)

Answer 4

\[\frac{m-\sqrt{n}}{r}\] must be root

Answer 5

I got it to \[(100x^4 +10x^2+1)(20x^2-2 +x)=0\]

Answer 6

Now I think we can solve it

Answer 7

200 :P

Answer 8

took me time to factorize :-/

Answer 9

Yes, cinar is right I got 200 as well

Answer 10

I love solving such questions, these are so fun. I hate school math

Answer 11

\[(ax^2+bx+c)*(dx^4+ex^3+fx^2+gx+h)=\]
=\[2000x^6+100x^3+x-2\]

Answer 12

Hmm factorization that's all, take some terms common and then try to take some more that's it

Answer 13

I like your way though

Answer 14

wow then, you saw it huh..

Answer 15

Did you use hit and trial. or some standard manipulatoin

Answer 16

cinar's way is a standard way.

mine is kind of hit and trial with inuition maybe

Answer 17

Fact about polynomials is that any polynomial can be factorised into quadratic and linear factors but it is not always feasible

Answer 18

Solivng a system of six equation doesnot seem promising also equatoins are not all linear

Answer 19

hmm yeah. so what do you suggest what should we use?

Answer 20

Cinar..we would find three sets of a,b,c
But how to solve the system

Answer 21

sorry guys, I have no time now, I can see it at night..

Answer 22

Alright Meet me tommorrow 12:00 pm IST

Answer 23

For more such problems Ishaan

Answer 24

IST? Indian Standard Time? you from India?

Answer 25

ya

Answer 26

Yeah I like these but it's late here and I have a problem to finish. I have posted it on OpenStudy you can check it out http://openstudy.com/users/ishaan94#/updates/4f2d8011e4b0571e9cba7c3b

Answer 27

Cool!