Jonask
For how many positive integer values of c does the equation below have an integer solution?
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Jonask
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\[\huge \color{brown}{ 2x^2+689x+c}\]
Jonask
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@TuringTest @phi @hartnn @sauravshakya @shubhamsrg
Jonask
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\[(2x+a)(x+b)=4x^2+2x(a+b)+ab\]
\[2(a+b)=689,ab=c\]
Jonask
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i dont kow wat to do here
TuringTest
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@JamesJ D:
Jonask
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first 689 can never be written as even number
TuringTest
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I am pretty bad at things like this for some reason, sorry.
Jonask
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thanks Turing
Jonask
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wolfram gives interger solutions as
\[c=689n-2n^2, x=-n\]
TuringTest
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I am assuming the equation is\[\large2x^2+689x+c=0\]right?
shubhamsrg
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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.
Jonask
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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
Jonask
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is this the only c
shubhamsrg
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What do you mean "the only c" ?
Jonask
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the question says
For how many positive integer values of c
so are there 344 interger c's
shubhamsrg
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yes.
Jonask
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okay thanks