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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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24 Aug 2016, 23:04
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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
[Reveal] Spoiler: OA

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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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.

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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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21 Sep 2017, 12:57
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Re: In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con   [#permalink] 21 Sep 2017, 12:57
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