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Jonask
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For how many positive integer values of c does the equation below have an integer solution?
 one year ago
 one year ago
Jonask Group Title
For how many positive integer values of c does the equation below have an integer solution?
 one year ago
 one year ago

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Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[\huge \color{brown}{ 2x^2+689x+c}\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
@TuringTest @phi @hartnn @sauravshakya @shubhamsrg
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[(2x+a)(x+b)=4x^2+2x(a+b)+ab\] \[2(a+b)=689,ab=c\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
i dont kow wat to do here
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
@JamesJ D:
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
first 689 can never be written as even number
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I am pretty bad at things like this for some reason, sorry.
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
thanks Turing
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=689
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
wolfram gives interger solutions as \[c=689n2n^2, x=n\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I am assuming the equation is\[\large2x^2+689x+c=0\]right?
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=2x%5E2%2B689x%2Bc%3D0
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
2x^2 + 689x + c = 0 c = 689x  2x^2 x= n will always given a ve c, hence x should be n , for some integer n, => c = 689n  2n^2 this has to be >0 => 689n 2n^2 >0 => n(689  2n) >0\ We already know n>0 hence we only have to deal with 689 > 2n or n < 344.5 Hence there can be 344 required value's of c. I have done this on brilliant.org if I am not wrong.
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
yes its from them,i have many other problems from them,if you dont mind you can check the question i asked bfore this one...thanks makes sense
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
is this the only c
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
What do you mean "the only c" ?
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
the question says For how many positive integer values of c so are there 344 interger c's
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
okay thanks
 one year ago
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