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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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Bunuel
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In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink] New post   09 Dec 2019, 01:13
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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


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gvij2017
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Re: In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink] New post   09 Dec 2019, 17:19
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x1*x2= 7,
x1+x2= -b

Statement 1
x can be only positive integer.
x can be 7, 1

7+1= 8= -b so b must be -8.
A is sufficient.

St. 2
b is negative.
Not sufficient.

A is answer.
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TestPrepUnlimited
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In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink] New post   Updated on: 04 Aug 2020, 19:41
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Bunuel 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


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Analyzing the Question:
If we have an equation in the form \(x^2 + ax + b= 0\), a is the negative sum of the two roots and b is the product of the two roots. So for \(x^2 + bx + 7= 0\) if we declare the two roots and m and n we have \(m*n = 7\) and \(m + n = -b\).

Statement 1:
m and n are positive and must be integers as x can only take integer values. Due to knowing \(m*n = 7\), we can only have m = 1, n = 7 or m = 7, n = 1. Thus either way we have \(b = -(m+n) = -8\)

Statement 2:
b can be any negative integer. Although the equation might not have solutions for a particular range of b, that is not an issue as we're asking for b. Insufficient.

Ans: A

Originally posted by TestPrepUnlimited on 09 Dec 2019, 09:46.
Last edited by TestPrepUnlimited on 04 Aug 2020, 19:41, edited 1 time in total.
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In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink] New post   09 Dec 2019, 13:31
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TestPrepUnlimited wrote:
Bunuel 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


Are You Up For the Challenge: 700 Level Questions


Analyzing the Question:
If we have an equation in the form \(x^2 + ax + b= 0\), a is the negative sum of the two roots and b is the product of the two roots. So for \(x^2 + bx + 7= 0\) if we declare the two roots and m and n we have \(m*n = 7\) and \(m + n = -b\).

Statement 1:
m and n are positive, therefore -b must be positive and b must be negative. Insufficient.

Statement 2:
Insufficient.

Combined:
We have duplicate information, we can automatically choose E. Besides m = 1 and n = 7 we can also have m = 0.5, n = 14, or m = n = sqrt(7) etc so there are multiple values of b we can take.

Statement A says postive integer value; can we use m = 0.5, n = 14 is it correct ? i think the answer is C
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In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink] New post   12 Jan 2020, 09:02
Statement 2 is insufficient because

roots can be 7 and 1 , which means b = -8 and roots can be 3.5 and 2 meaning that b = -5.5.

Hence, insufficient.
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] New post   04 Aug 2020, 09:40
gvij2017 wrote:
x1*x2= 7,
x1+x2= -b

Statement 1
x can be only positive integer.
x can be 7, 1

7+1= 8= -b so b must be -8.
A is sufficient.

St. 2
b is negative.
Not sufficient.

A is answer.

Why can x only be 7 or 1? It isn't mentioned that x can take only integral values.
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Re: In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink] New post   04 Aug 2020, 13:25
ShankSouljaBoi wrote:
Statement 2 is insufficient because

roots can be 7 and 1 , which means b = -8 and roots can be 3.5 and 2 meaning that b = -5.5.

Hence, insufficient.
A

b is given as a negative integer. but in your example it’s not an integer.

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In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink] New post   04 Aug 2020, 13:43
TestPrepUnlimited
TestPrepUnlimited wrote:
Bunuel 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


Are You Up For the Challenge: 700 Level Questions


Analyzing the Question:
If we have an equation in the form \(x^2 + ax + b= 0\), a is the negative sum of the two roots and b is the product of the two roots. So for \(x^2 + bx + 7= 0\) if we declare the two roots and m and n we have \(m*n = 7\) and \(m + n = -b\).

Statement 1:
m and n are positive, therefore -b must be positive and b must be negative. Insufficient.

Statement 2:
Insufficient.

Combined:
We have duplicate information, we can automatically choose E. Besides m = 1 and n = 7 we can also have m = 0.5, n = 14, or m = n = sqrt(7) etc so there are multiple values of b we can take.

Can you please elaborate how statement B is insufficient. Maybe a counterexample.
Thanks in advance :)
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Re: In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink] New post   04 Aug 2020, 19:45
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Hi AdiBir ,

Thanks for the question! I edited my original post, we're asking for b and the given equation doesn't tell us anything about b. Then with only statement 2, b is allowed to be any negative integer and since we can't confirm a single value it would be insufficient.
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In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink] New post   21 Mar 2022, 10:19
ShankSouljaBoi wrote:
Statement 2 is insufficient because

roots can be 7 and 1 , which means b = -8 and roots can be 3.5 and 2 meaning that b = -5.5.

Hence, insufficient.
A



But b is a negative integer. And an integer plus a fraction can't be an integer.

However, there can be more than two set of fractional divisors of 7, which give a negative number when added.

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In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con [#permalink]
21 Mar 2022, 10:19
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