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

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

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

MBM

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Source: Beat The GMAT — Problem Solving | Tue May 26, 2009 5:04 am

dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

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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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

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).

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
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dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

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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