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There are exactly N distinct rational numbers k such that |k| < 200 an

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There are exactly N distinct rational numbers k such that |k| < 200 an  [#permalink]

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New post 19 Mar 2019, 09:11
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There are exactly N distinct rational numbers k such that \(|k| < 200\) and \(5x^2 + kx + 12 = 0\)
has at least one integer solution for x. What is N?

(A) 6

(B) 12

(C) 24

(D) 48

(E) 78

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Re: There are exactly N distinct rational numbers k such that |k| < 200 an  [#permalink]

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New post 20 Mar 2019, 19:53
Noshad wrote:
There are exactly N distinct rational numbers k such that \(|k| < 200\) and \(5x^2 + kx + 12 = 0\)
has at least one integer solution for x. What is N?

(A) 6

(B) 12

(C) 24

(D) 48

(E) 78



Let us take k to one side => \(5x^2 + kx + 12 = 0........k=-5x-\frac{12}{x}............\)
Now |k|<200 or -200<k<200, that is \(-200<-5x-\frac{12}{x}<200\)
-5x will become equal to 200, the moment x is 40, so x has to be less than 40..
Similarly -5x will become greater than 200, the moment x is -40, so x has to be greater than -40..
Thus x are from -39 to 39, which equals 2*39+1=79, which includes k as 0.
However k as 0 is not possible as the equation is not defined at that point =>\(5x^2 + kx + 12 = 0..........5x^2+12=0.....5x^2=-12\) NOT possible as \(5x^2\geq{0}\)
Thus 79-1=78

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Re: There are exactly N distinct rational numbers k such that |k| < 200 an  [#permalink]

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New post 20 Mar 2019, 20:09
chetan2u wrote:
Noshad wrote:
There are exactly N distinct rational numbers k such that \(|k| < 200\) and \(5x^2 + kx + 12 = 0\)
has at least one integer solution for x. What is N?

(A) 6

(B) 12

(C) 24

(D) 48

(E) 78



Let us take k to one side => \(5x^2 + kx + 12 = 0........k=-5x-\frac{12}{x}............\)
Now |k|<200 or -200<k<200, that is \(-200<-5x-\frac{12}{x}<200\)
-5x will become equal to 200, the moment x is 40, so x has to be less than 40..
Similarly -5x will become greater than 200, the moment x is -40, so x has to be greater than -40..

Thus x are from -39 to 39, which equals 2*39+1=79, which includes k as 0.
However k as 0 is not possible as the equation is not defined at that point =>\(5x^2 + kx + 12 = 0..........5x^2+12=0.....5x^2=-12\) NOT possible as \(5x^2\geq{0}\)
Thus 79-1=78

E


Chetan2u can you explain the highlighted part.
why does -5x becoming 400 is a problem. Since K will still b -5x-12/x which is less than 400
Similarly for next stmt
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Re: There are exactly N distinct rational numbers k such that |k| < 200 an  [#permalink]

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New post 20 Mar 2019, 20:43
globaldesi wrote:
chetan2u wrote:
Noshad wrote:
There are exactly N distinct rational numbers k such that \(|k| < 200\) and \(5x^2 + kx + 12 = 0\)
has at least one integer solution for x. What is N?

(A) 6

(B) 12

(C) 24

(D) 48

(E) 78



Let us take k to one side => \(5x^2 + kx + 12 = 0........k=-5x-\frac{12}{x}............\)
Now |k|<200 or -200<k<200, that is \(-200<-5x-\frac{12}{x}<200\)
-5x will become equal to 200, the moment x is 40, so x has to be less than 40..
Similarly -5x will become greater than 200, the moment x is -40, so x has to be greater than -40..

Thus x are from -39 to 39, which equals 2*39+1=79, which includes k as 0.
However k as 0 is not possible as the equation is not defined at that point =>\(5x^2 + kx + 12 = 0..........5x^2+12=0.....5x^2=-12\) NOT possible as \(5x^2\geq{0}\)
Thus 79-1=78

E


Chetan2u can you explain the highlighted part.
why does -5x becoming 400 is a problem. Since K will still b -5x-12/x which is less than 400
Similarly for next stmt


-5x will become 200 when x is -40, but -12/x will also become positive (-12)/(-40)=12/40, so sum is 200+(12/40) >200
Similarly for the other case..
-12/x will have SAME sign as -5x, so will get added, either both as negative, then <-200 or both as positive, then >200
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Re: There are exactly N distinct rational numbers k such that |k| < 200 an   [#permalink] 20 Mar 2019, 20:43
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