The solution to x ^2 + 7 x + 12 > 0 is
A. -4 < x < -3
B. x < -4 and x > -3
C. x > -4
D. x < -3
E. 7 < x < 12
MBM
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solution to x ^2 + 7 x + 12 > 0
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sanju09
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Sanjeev K Saxena
Quantitative Instructor
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www.manyagroup.com
Find the zero's of the inequality which will be excluded and plot them on a number line, or imagine them on a number line. You will have 3 intervalssanju09 wrote:The solution to x ^2 + 7 x + 12 > 0 is
A. -4 < x < -3
B. x < -4 and x > -3
C. x > -4
D. x < -3
E. 7 < x < 12
MBM
(x+4)(x+3)>0
zero's are -4, -3.
Intervals
x< -4, -4<x<-3, x>-3
The middle interval fails-- the product is negative in this region for any x- and the union of the first and third gives the answer.
Hence B.
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Stuart@KaplanGMAT
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Slightly different approach (but very similar):dtweah wrote:Find the zero's of the inequality which will be excluded and plot them on a number line, or imagine them on a number line. You will have 3 intervalssanju09 wrote:The solution to x ^2 + 7 x + 12 > 0 is
A. -4 < x < -3
B. x < -4 and x > -3
C. x > -4
D. x < -3
E. 7 < x < 12
MBM
(x+4)(x+3)>0
zero's are -4, -3.
Intervals
x< -4, -4<x<-3, x>-3
The middle interval fails-- the product is negative in this region for any x- and the union of the first and third gives the answer.
Hence B.
To get a positive product, both brackets must have the same sign.
Therefore, either:
x + 4 > 0 and x + 3 > 0
or
x + 4 < 0 and x + 3 < 0
In the first case, we get:
x > -4 AND x > -3. We always take the more restrictive condition:
x > -3
In the second case, we get
x < -4 and x < -3
Again, taking the more restrictive condition:
x < -4
So, x < -4 OR x > -3.. choose (b).
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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This used to be my favorite way of tackling these but I switched to the sign method.Stuart Kovinsky wrote:Slightly different approach (but very similar):dtweah wrote:Find the zero's of the inequality which will be excluded and plot them on a number line, or imagine them on a number line. You will have 3 intervalssanju09 wrote:The solution to x ^2 + 7 x + 12 > 0 is
A. -4 < x < -3
B. x < -4 and x > -3
C. x > -4
D. x < -3
E. 7 < x < 12
MBM
(x+4)(x+3)>0
zero's are -4, -3.
Intervals
x< -4, -4<x<-3, x>-3
The middle interval fails-- the product is negative in this region for any x- and the union of the first and third gives the answer.
Hence B.
To get a positive product, both brackets must have the same sign.
Therefore, either:
x + 4 > 0 and x + 3 > 0
or
x + 4 < 0 and x + 3 < 0
In the first case, we get:
x > -4 AND x > -3. We always take the more restrictive condition:
x > -3
In the second case, we get
x < -4 and x < -3
Again, taking the more restrictive condition:
x < -4
So, x < -4 OR x > -3.. choose (b).
