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anonymous
 4 years ago
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
anonymous
 4 years ago
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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{m\sqrt{n}}{r}\] must be root

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got it to \[(100x^4 +10x^2+1)(20x^22 +x)=0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now I think we can solve it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0took me time to factorize :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, cinar is right I got 200 as well

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I love solving such questions, these are so fun. I hate school math

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(ax^2+bx+c)*(dx^4+ex^3+fx^2+gx+h)=\] =\[2000x^6+100x^3+x2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm factorization that's all, take some terms common and then try to take some more that's it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I like your way though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wow then, you saw it huh..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Did you use hit and trial. or some standard manipulatoin

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cinar's way is a standard way. mine is kind of hit and trial with inuition maybe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Fact about polynomials is that any polynomial can be factorised into quadratic and linear factors but it is not always feasible

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Solivng a system of six equation doesnot seem promising also equatoins are not all linear

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm yeah. so what do you suggest what should we use?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Cinar..we would find three sets of a,b,c But how to solve the system

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry guys, I have no time now, I can see it at night..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Alright Meet me tommorrow 12:00 pm IST

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For more such problems Ishaan

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0IST? Indian Standard Time? you from India?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah 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
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