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Retired Moderator
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(x - 3) / (x^2 + 3x + 2) ≥ 1
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24 Sep 2021, 10:39
24 Sep 2021, 10:39
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\(\frac{(x - 3)}{(x^2 + 3x + 2)} ≥ 1\)
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Number of integers which satisfy the above inequality |
1 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
(x - 3) / (x^2 + 3x + 2) ≥ 1
[#permalink]
24 Sep 2021, 15:04
24 Sep 2021, 15:04
1
let's dissect denominator=0 constraint firstly, \(x^2+3x+2=0\), \((x+1)(x+2)=0\) and x cannot be -1 and/or -2
next, both numerator and denominator must be either positive of negative
we safely multiply as signs of both sides are the same, and it leads to 2 inequalities that follow
\(x-3≥x^2+3x+2\) and \(x^2+3x+2≥x-3\)
\(x^2+2x+5≤0\) and \(x^2+2x+5≥0\) together these satisfy the equation \(x^2+2x+5=0\)
by checking the discriminant of the latter quadratic equation, \(2^2-4*5*1=-16\) we see that it's negative ... and there is not any solution
answer is \(D\)
next, both numerator and denominator must be either positive of negative
we safely multiply as signs of both sides are the same, and it leads to 2 inequalities that follow
\(x-3≥x^2+3x+2\) and \(x^2+3x+2≥x-3\)
\(x^2+2x+5≤0\) and \(x^2+2x+5≥0\) together these satisfy the equation \(x^2+2x+5=0\)
by checking the discriminant of the latter quadratic equation, \(2^2-4*5*1=-16\) we see that it's negative ... and there is not any solution
answer is \(D\)
KarunMendiratta wrote:
\(\frac{(x - 3)}{(x^2 + 3x + 2)} ≥ 1\)
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Number of integers which satisfy the above inequality |
1 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
General Discussion
Re: (x - 3) / (x^2 + 3x + 2) ≥ 1
[#permalink]
24 Sep 2021, 21:58
24 Sep 2021, 21:58
motion2020 wrote:
let's dissect denominator=0 constraint firstly, \(x^2+3x+2=0\), \((x+1)(x+2)=0\) and x cannot be -1 and/or -2
next, both numerator and denominator must be either positive of negative
we safely multiply as signs of both sides are the same, and it leads to 2 inequalities that follow
\(x-3≥x^2+3x+2\) and \(x^2+3x+2≥x-3\)
\(x^2+2x+5≤0\) and \(x^2+2x+5≥0\) together these satisfy the equation \(x^2+2x+5=0\)
by checking the discriminant of the latter quadratic equation, \(2^2-4*5*1=-16\) we see that it's negative ... and there is not any solution
answer is \(D\)
interesting problem. More likely to be found in CAT exam.
next, both numerator and denominator must be either positive of negative
we safely multiply as signs of both sides are the same, and it leads to 2 inequalities that follow
\(x-3≥x^2+3x+2\) and \(x^2+3x+2≥x-3\)
\(x^2+2x+5≤0\) and \(x^2+2x+5≥0\) together these satisfy the equation \(x^2+2x+5=0\)
by checking the discriminant of the latter quadratic equation, \(2^2-4*5*1=-16\) we see that it's negative ... and there is not any solution
answer is \(D\)
interesting problem. More likely to be found in CAT exam.
KarunMendiratta wrote:
\(\frac{(x - 3)}{(x^2 + 3x + 2)} ≥ 1\)
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Number of integers which satisfy the above inequality |
1 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
