Question

For how many positive integer values of c does the equation below have an integer solution?

Answer 1

\[\huge \color{brown}{ 2x^2+689x+c}\]

Answer 2

@TuringTest @phi @hartnn @sauravshakya @shubhamsrg

Answer 3

\[(2x+a)(x+b)=4x^2+2x(a+b)+ab\]
\[2(a+b)=689,ab=c\]

Answer 4

i dont kow wat to do here

Answer 5

@JamesJ D:

Answer 6

first 689 can never be written as even number

Answer 7

I am pretty bad at things like this for some reason, sorry.

Answer 8

thanks Turing

Answer 9

http://www.wolframalpha.com/input/?i=689

Answer 10

wolfram gives interger solutions as
\[c=689n-2n^2, x=-n\]

Answer 11

I am assuming the equation is\[\large2x^2+689x+c=0\]right?

Answer 12

http://www.wolframalpha.com/input/?i=2x%5E2%2B689x%2Bc%3D0

Answer 13

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.

Answer 14

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

Answer 15

is this the only c

Answer 16

What do you mean "the only c" ?

Answer 17

the question says
For how many positive integer values of c
so are there 344 interger c's

Answer 18

yes.

Answer 19

okay thanks