Question on inequality : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 18 Jan 2017, 21:37

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Question on inequality

Author Message
Director
Joined: 28 Dec 2005
Posts: 755
Followers: 1

Kudos [?]: 14 [0], given: 0

### Show Tags

25 Jun 2006, 15:54
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can someone help me solve for x?

(3x +2)/(x^2 + x -6) <= 1

I know how to get to a certain point, i.e. find the values of x for which the numerator and denominator become 0, but how do I find the range of values for x based on that?[/code]
Manager
Joined: 09 Apr 2006
Posts: 173
Location: Somewhere in Wisconsin!
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

25 Jun 2006, 16:19
The expression reduces to x^^2+2x-8>=0
This could be further reduced to (x - 4)(x + 2) >=0

Since it is a inequality please check both the expressions in the parantheses. They can be both greater than zero or both less than zero.

You should get x >=4 or x <=-2.
_________________

Thanks,
Zooroopa

Director
Joined: 28 Dec 2005
Posts: 755
Followers: 1

Kudos [?]: 14 [0], given: 0

### Show Tags

25 Jun 2006, 16:42
Zooroopa wrote:
The expression reduces to x^^2+2x-8>=0
This could be further reduced to (x - 4)(x + 2) >=0

Since it is a inequality please check both the expressions in the parantheses. They can be both greater than zero or both less than zero.

You should get x >=4 or x <=-2.

With all due respect, this approach seems flawed. You would need to consider the denominator as well, and values for which it becomes 0. The equation actually reduces to {[(x-4)(x+2)]/[(x+3)(x-2)]} >= 0. I know from this that we can get ranges of x.
Manager
Joined: 09 Apr 2006
Posts: 173
Location: Somewhere in Wisconsin!
Followers: 1

Kudos [?]: 3 [0], given: 0

### Show Tags

25 Jun 2006, 21:00
Futuristic wrote:
Zooroopa wrote:
The expression reduces to x^^2+2x-8>=0
This could be further reduced to (x - 4)(x + 2) >=0

Since it is a inequality please check both the expressions in the parantheses. They can be both greater than zero or both less than zero.

You should get x >=4 or x <=-2.

With all due respect, this approach seems flawed. You would need to consider the denominator as well, and values for which it becomes 0. The equation actually reduces to {[(x-4)(x+2)]/[(x+3)(x-2)]} >= 0. I know from this that we can get ranges of x.

Respectfully noted the point. x = 2 is not in the range.

However how are you resolving the numerator into the expression in bold?
_________________

Thanks,
Zooroopa

Director
Joined: 28 Dec 2005
Posts: 755
Followers: 1

Kudos [?]: 14 [0], given: 0

### Show Tags

28 Jun 2006, 13:16
VP
Joined: 21 Sep 2003
Posts: 1065
Location: USA
Followers: 3

Kudos [?]: 73 [0], given: 0

### Show Tags

28 Jun 2006, 14:37
Futuristic wrote:
Zooroopa wrote:
The expression reduces to x^^2+2x-8>=0
This could be further reduced to (x - 4)(x + 2) >=0

Since it is a inequality please check both the expressions in the parantheses. They can be both greater than zero or both less than zero.

You should get x >=4 or x <=-2.

With all due respect, this approach seems flawed. You would need to consider the denominator as well, and values for which it becomes 0. The equation actually reduces to {[(x-4)(x+2)]/[(x+3)(x-2)]} >= 0. I know from this that we can get ranges of x.

{[(x-4)(x+2)]/[(x+3)(x-2)]} >= 0
This basically means that we have the following ranges of values for x
(-infinity, -3) (-3,-2) (-2,2) (2,4) (4, infinity)

Now pick a vaue from each range and verify if the expression makes sense:
For eg: Range (-infinity, -3). Pick x = -4
(-4-4)(-4+2)/(-4+3)(-4-2) = (-16)*(-2)/(-1)*(2) = -16 which <=0. Hence
x has values in the range to (-infinity, -3) or x < -3

Similarly you can verify for the other 4 ranges. You will see that some of them don't satisfy the inequality.
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

Manager
Joined: 26 Jun 2006
Posts: 152
Followers: 1

Kudos [?]: 6 [0], given: 0

### Show Tags

28 Jun 2006, 14:55
Let me try. From {[(x-4)(x+2)]/[(x+3)(x-2)]} >= 0 we consider 2 options: the inequality holds if either 1) Num>=0 and Den>0 or 2) Num<=0 and Den<0. So,

1. (x-4)(x+2)>=0 and (x+3)(x-2)>0
or x in (-infty,-3) or [4,+infty)

2. (x-4)(x+2)<=0 and (x+3)(x-2)<0,
in which case x in (2,2]

Combining 1 and 2, the answer is x in (-infty,-3) or (2,2] or [4,+infty)

Too much calculations to my taste...
28 Jun 2006, 14:55
Display posts from previous: Sort by