quadratic inequalities

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by ronniecoleman » Sat Jan 10, 2009 9:17 am
IMO B


X^2- 4X + 3 < 0


<1 X < 3

Since x is integer..so x is 2
suff


X^2+ 4X+ 3 > 0

x < -3

x > -1

Insuff
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by Brent@GMATPrepNow » Sat Jan 10, 2009 9:29 am
Nice work, Ronnie.
Here's my full solution as well:

(1) To solve quadratic inequalities, try factoring the expressions.
We get (x+1)(x+3) > 0 Notice that when x=-1 and x=-3, (x+1)(x+3) EQUALS 0.
Now ask, “What happens when x is less than -3?” We see that (x+1)(x+3) will always be positive for these values of x.
Then ask, “What happens when x is between -3 and -1?” We see that (x+1)(x+3) will always be negative for these values of x.
Finally ask, “What happens when x is greater than -1?” We see that (x+1)(x+3) will always be positive for these values of x.
We see that (1) tells us that x can be less than -3 or greater than -1 (not enough information)

(2) Using the same technique as above, we see that -3 > x > -1
Since x is an integer, it must be the case that x= -2 (enough information)

The correct answer is B
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by cramya » Sat Jan 10, 2009 11:49 am
Using the same technique as above, we see that -3 > x > -1
Brent,

I am sure u meant -1>x>-3

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by DanaJ » Sat Jan 10, 2009 12:28 pm
I would eliminate A&D in a snap: all you have to notice is that every positive integer has x^2+4x+3>0.
Then you would have x^2-4x+3 = (x-1)(x-3). This equation is negative only between the roots, which means that 1 < x < 3. And there's only one integer who fits that description: 2.

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by logitech » Sat Jan 10, 2009 12:46 pm
DanaJ wrote:I would eliminate A&D in a snap: all you have to notice is that every positive integer has x^2+4x+3>0.
Then you would have x^2-4x+3 = (x-1)(x-3). This equation is negative only between the roots, which means that 1 < x < 3. And there's only one integer who fits that description: 2.
This is how you attack a DS problem like a shark. Great work Danaj!
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by aj5105 » Sun Jan 11, 2009 3:35 am
(B)