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anonymous
 5 years ago
Hi everybody! Find all positive integral solutions (x,y) of the polynomial equation:
4x^3+4x^2y15xy^218y^312x^2+6xy+36y^2+5x10y=0
anonymous
 5 years ago
Hi everybody! Find all positive integral solutions (x,y) of the polynomial equation: 4x^3+4x^2y15xy^218y^312x^2+6xy+36y^2+5x10y=0

This Question is Closed

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0couldn't u find a longer eq? :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its a monstrous beast...

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0I think it is smth like (axby)^3 bu i can't tell wht r a and b

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good try angela..but u never know that will factorize completely to 3 linear factors!... Infact this BEAST can be factored to any one of the following: 1.3 linears. 2.1 linear and 1 conic

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(x2*y)*(2*x1+3*y)*(2*x5+3*y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(x2y)(2x+3y5)(2x+3y1)=0

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0i'm not planning to deal with this..u do it...:P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got it first, please medal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u rock cantorset! but we cant use comps!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you know i use computer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@contorset: lol, its obvious. :P

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0hey how did u do tht?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well witout at computer its LUCK..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I didn't do it on my own, wolframalpha

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0and how do u do it in comp?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha... if u had a good luck u wud put x=2y and inspect!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can do it in wolframalpha,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then u get (x2y) and get others by dividing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i should not give out my secrets though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k. but tat doesnt solve the problem..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x=2y ? you mean substitute

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@saubhik: You need a VERY good luck for that!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but see tat doesnt..solve the problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have to find positive integer solutions..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but you can pick anything, x = 4y for instance

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0and i thought u guys were smart :P:P:P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont see any logic in this problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ya cantorset that's truly possible but as i said we need luck and brainbreaking insights...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see..now how to find the solution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the first one(x2y), x = 2I y=I

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u can have infinite pairs (2n,n) for a positive integer n since x2y is a factor...what for the other two?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its obvious (1,1) but are there others..THIS IS DIOPHANTINE!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK lets see, x + 3y/2 =5/2 x=(53y)/2 53y should be even,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what about 2x+3y=1? yay...theres no positive integers here in this equation..so this is infertile:P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how can u get postive integers (x,y) from 2x+3y<5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0infertile..mother has no babies:P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that can be solved with euclids algorithm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh sorry, i dint read that 'positive' part! :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this mother 2x+3y=5 has only one baby i.e. (1,1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So our answer: (1,1) and (2n,n)

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0OMG!!! This is giving me headache only by reading ur explanations imagine if i had to solve this....:@@@@@@@@

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So its done, (n,2n) U (1,1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Openstudy rocks!! thnx evrybody!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oops (2n,n). yes right

angela210793
 5 years ago
Best ResponseYou've already chosen the best response.0Going to deal with literature bye and have a good time :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0, amog, you wrote x= 1, y = 21?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@angela: buhbye. @cantorset: It was an 'I' (eye)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the first one(x2y), x = 2I y=I

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes i corrected it, didn't I?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hi cantorset...do u mind givin me an idea how to have a factor to the eqn..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0suppose i have the baddest luck and go on inspecting from x=3y;x=4y....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, ... start over . why would i think to use x = 3y or anything like that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also its 3 linear, not a conic

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or worser x=3y+7 etc..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you mean because the solution is ( ax + by ) ( dx + cy + ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0See cantormath..the given equation Infact this BEAST can be factored to any one of the following: 1.3 linears. 2.1 linear and 1 conic

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now clearly theres no constant term in this equation...so a linear factor (which must be present) must pass through the origin...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and so we start checking by x=y;x=2y;x=3y...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if the question does not mean to torment us we can get the factor in a few trials...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k..thnx evrybody for helping..:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, youre saying in general, a cubic in x and y can be factored into either 3 linears, or 1 linear and 2 conics? , or 2 linears, and a conic?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you mean by constant term in the equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0term with no variables attched..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the final solution is 3 linears, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yups...family of 3 straight line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool, so ok , theres no constant term, and clearly, (0,0) works

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yups..try looking for x=y,2y,3y,...here u need luck

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so that tells us that x = ay or y = ax must be one of the solutions?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, but thats only if you do the long division< right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you know to assume, 2x = y ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or wait, you know that (0,0) is a solution and (1,1) is a solution, right>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes...not assume but see if its correct by replacing 2x by y OR replacing y by 2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(1,1) is known after we get other factors using x2y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm, yes but i mean is there something that would make us want to to do x = 2y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0( 1,1) is another solution then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OR if u knew the problem beforehand OR if u have wolframalpha open on ur browser. No...1 is not a POSITIVE integer :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04x^2 + 9y^2 + 12xy 2x 15y +5 =0 How do you solve this btw?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0from discriminant criteria for various conics see if its parabola/ellipse/hyperbola..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k. for positive integer solutions...ALL 2VARIABLE EQUATION DO NOT HAVE positive integer solutions.....such questions are given after testing them..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So here its a parabola

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tats what wolframalpha says...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I didn't use wolframalpha for it? Afterall we in India are not allowed to use any calc! :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what are the possible shapes of a third degree x,y equation, a conic is second degree,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Cool, did you give the JEE?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0may be: 1.family of straight lines.. 2. a strt line and a conic

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gave it! but didnt get it:(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks :) So what plans?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0very cool saubhik, didnt know that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont know ....awaiting for AIEEE and WBJEE and ISI results...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0All India Engineering Entrance Examination (in india)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0saubhik, i got a problem for ya

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mayb ei should post new one
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