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solution to x ^2 + 7 x + 12 > 0

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solution to x ^2 + 7 x + 12 > 0

by sanju09 » Tue May 26, 2009 5:04 am
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



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Re: solution to x ^2 + 7 x + 12 > 0

by dtweah » Tue May 26, 2009 5:15 am
sanju09 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
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 intervals
(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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Re: solution to x ^2 + 7 x + 12 > 0

by Stuart@KaplanGMAT » Tue May 26, 2009 9:57 am
dtweah wrote:
sanju09 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
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 intervals
(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.
Slightly different approach (but very similar):

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).
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Re: solution to x ^2 + 7 x + 12 > 0

by dtweah » Tue May 26, 2009 12:26 pm
Stuart Kovinsky wrote:
dtweah wrote:
sanju09 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
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 intervals
(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.
Slightly different approach (but very similar):

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).
This used to be my favorite way of tackling these but I switched to the sign method.
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