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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 09 Dec 2019, 00:13
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A
B
C
D
E

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(N/A)

Question Stats:

17% (01:41) correct 83% (02:00) wrong based on 53 sessions

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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 09 Dec 2019, 08:46
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, 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.
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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 09 Dec 2019, 12:31
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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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 09 Dec 2019, 16:19
1
1
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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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 12 Jan 2020, 08:02
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.
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In the equation x^2 + bx + 7= 0, where x is a variable, and b is a con   [#permalink] 12 Jan 2020, 08:02
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