If x is an integer, what is the value of x? (1) x^2 + 4x + 3 > 0 (2) x^2 – 4x + 3 < 0

IMO B X^2- 4X + 3 0 x -1 Insuff

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

Using the same technique as above, we see that -3 > x > -1 Brent, I am sure u meant -1>x>-3 Regards, CR

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.

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!