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Question on inequality

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Question on inequality [#permalink] New post 25 Jun 2006, 15:54
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]
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 [#permalink] New post 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.
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 [#permalink] New post 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.
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 [#permalink] New post 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?
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 [#permalink] New post 28 Jun 2006, 13:16
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 [#permalink] New post 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.


Your approach is correct.
{[(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.
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 [#permalink] New post 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...
  [#permalink] 28 Jun 2006, 14:55
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