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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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Intern
Joined: 14 Mar 2016
Posts: 1
In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con  [#permalink]

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Updated on: 18 Sep 2016, 09:28
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Difficulty:

95% (hard)

Question Stats:

36% (01:53) correct 64% (01:47) wrong based on 76 sessions

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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, 22:04.
Last edited by Bunuel on 18 Sep 2016, 09:28, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Senior Manager
Joined: 23 Apr 2015
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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, 22:29
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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.

+1 for kudos
##### General Discussion
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Joined: 11 Mar 2018
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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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10 Jun 2018, 10:29
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.

+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 &nbs [#permalink] 10 Jun 2018, 10:29
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