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In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con

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In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink]

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New post Updated on: 18 Sep 2016, 10:28
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In the equation x^2 + bx + 7= 0, where x is a variable, and b is a constant. What is the value of b?

(1) x can take only positive integer values
(2) b is a negative integer

Originally posted by AnanyaSachdev on 24 Aug 2016, 23:04.
Last edited by Bunuel on 18 Sep 2016, 10:28, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink]

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New post 24 Aug 2016, 23:29
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AnanyaSachdev wrote:
In the equation x^2 + bx + 7= 0, where x is a variable, and b is a constant. What is the value of b?
1) x can take only positive integer values
2) b is a negative integer



Given \(x^2 + bx + 7= 0\) and find \(b\) ?
Consider 1) \(x\) can take only positive integer values : This means the equation can be of the form \((x - m) * (x - n)\) where \(m\) and \(n\) are positive integer roots. The resulting equation will be of the form \(x^2 - x(m+n) + m*n = 0 = x^2 + bx + 7\). So \(m*n = 7\) and since \(7\) is prime, the only decomposition is \(1\) and \(7\), giving \(b = -(m+n) = -8\).
Sufficient,
Hence A or D

Consider 2) \(b\) is a negative integer.
\(b\) can any value giving roots as non-integers. Not sufficient.

Answer is A.



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Re: In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink]

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New post 10 Jun 2018, 11:29
Senthil1981 wrote:
AnanyaSachdev wrote:
In the equation x^2 + bx + 7= 0, where x is a variable, and b is a constant. What is the value of b?
1) x can take only positive integer values
2) b is a negative integer



Given \(x^2 + bx + 7= 0\) and find \(b\) ?
Consider 1) \(x\) can take only positive integer values : This means the equation can be of the form \((x - m) * (x - n)\) where \(m\) and \(n\) are positive integer roots. The resulting equation will be of the form \(x^2 - x(m+n) + m*n = 0 = x^2 + bx + 7\). So \(m*n = 7\) and since \(7\) is prime, the only decomposition is \(1\) and \(7\), giving \(b = -(m+n) = -8\).
Sufficient,
Hence A or D

Consider 2) \(b\) is a negative integer.
\(b\) can any value giving roots as non-integers. Not sufficient.

Answer is A.



+1 for kudos


The answer to 1 seems clear, but I am confused about the answer to 2 and would appreciate insight..

\(b<0\) and is an integer.

Using your example for 1, if \(m*n = 7\), and add to \(b\)then both \(m\) and \(n\) are negative integers.

\(b = -7-1 = -8\)

Thanks.
Re: In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con   [#permalink] 10 Jun 2018, 11:29
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In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con

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