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x^2+8x+7<0...................basic stuff but intriguining.

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x^2+8x+7<0...................basic stuff but intriguining. [#permalink] New post 25 Nov 2012, 00:58
Hi,

My first post on GMAT Club.I have just kick started my prep for GMAT, get always stuck at the inequalities. Need a bit of help.

Q. x^2+8x+7<0

solution is -7<x<-1

i can break down the equation to (x+1)(x+7)<0

post this x+1<0 the x<-1 which is true as per the answer
but if i follow the similar process for x+7<0 the x<-7 which is not true as per the answer.

please clear my basic, when "<" sign is there in question does the inequality equation follow certain rules.

one more similar question

Q x^2+7x+12>0

answer: x<-4 and x>-3

according to me it should be x>-3 and x>-4 hence solution should be x>-4 .................

its confusing :? .............please help!!!!!!
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1 KUDOS received
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Status: Done with formalities.. and back..
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Re: x^2+8x+7<0...................basic stuff but intriguining. [#permalink] New post 25 Nov 2012, 01:07
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victorywillbemine wrote:
Hi,

My first post on GMAT Club.I have just kick started my prep for GMAT, get always stuck at the inequalities. Need a bit of help.

Q. x^2+8x+7<0
solution is -7<x<-1
i can break down the equation to (x+1)(x+7)<0
post this x+1<0 the x<-1 which is true as per the answer
but if i follow the similar process for x+7<0 the x<-7 which is not true as per the answer.
please clear my basic, when "<" sign is there in question does the inequality equation follow certain rules.
one more similar question
Q x^2+7x+12>0
answer: x<-4 and x>-3
according to me it should be x>-3 and x>-4 hence solution should be x>-4 .................
its confusing :? .............please help!!!!!!

Welcome to GMAT club.

The rule is once u get the solution you need to ascertain its validity or you need to plug in that solution back in equation and test for validity. for example:

for first equation, once you've reached to (x+1)(x+7)<0
Now you know that for this to be true, one term must be negative and other positive.
say (x+1) is negative => x <-1
but at this time x+7 must be positive, so x+7 > 0 => x>-7
Hence we get the range -7 <x<-1

Again second possibility,
if x+1 is positive then x+7 should be negative or
x+1 >0 => x >-1
and x+7 <0 => x < -7
but it is not possible. Since x can not be greater than -1 and less than -7 at same time.

So you get that the solution obtained first was correct one. and range is -7<x<-1
similarly you can solve the second problem.

Once you are comfortble getting ans like this, then slowly u can move to graphical approach which makes it even easier to get the answer by knowing the concept.

Hope it helps.
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Re: x^2+8x+7<0...................basic stuff but intriguining. [#permalink] New post 25 Nov 2012, 02:22
My very first reply on the post............yippie :lol:

You cleared the funda completely............. My first kudo to you!!!!!! :-D
but just to throw more light on the concept. Do you have material on board to supplement the solution more.

For eg if instead of two cases we have three, how shall I go about in both less than and greater than signs
Q. x(x+1)(x+2)<0
Q. x(x+1)(x+2)>0
Q. (x+1)(x+2)(x+3)<0
Q. (x+1)(x+2)(x+3)>0
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Re: x^2+8x+7<0...................basic stuff but intriguining. [#permalink] New post 25 Nov 2012, 02:29
victorywillbemine wrote:
My very first reply on the post............yippie :lol:

You cleared the funda completely............. My first kudo to you!!!!!! :-D
but just to throw more light on the concept. Do you have material on board to supplement the solution more.

For eg if instead of two cases we have three, how shall I go about in both less than and greater than signs
Q. x(x+1)(x+2)<0
Q. x(x+1)(x+2)>0
Q. (x+1)(x+2)(x+3)<0
Q. (x+1)(x+2)(x+3)>0

Ha ha.. i love kudos :-D

Yup, graphical approach... Search for inequalities and you could probably stumble upon the theory. I can give u the concept (not the theory) and way to solve in brief.

take any of the examples, say 1.
x(x+1)(x+2)<0

we have 3 decision points. at 0, -1, and -2 clearly from equation.
arrange these numbers in ascending order.
-2 -1 0
Start with P from right and keep adding P and N alternatively.
you get:
N -2 P -1 N 0 P
Here, N and P defines the range for you where ineuality is positive and negative.
For your question inequality is negative.

therefore the range for x can be -1<x <0 or x<-2

Hope it helps.
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Re: x^2+8x+7<0...................basic stuff but intriguining. [#permalink] New post 25 Nov 2012, 04:38
Expert's post
victorywillbemine wrote:
Hi,

My first post on GMAT Club.I have just kick started my prep for GMAT, get always stuck at the inequalities. Need a bit of help.

Q. x^2+8x+7<0

solution is -7<x<-1

i can break down the equation to (x+1)(x+7)<0

post this x+1<0 the x<-1 which is true as per the answer
but if i follow the similar process for x+7<0 the x<-7 which is not true as per the answer.

please clear my basic, when "<" sign is there in question does the inequality equation follow certain rules.

one more similar question

Q x^2+7x+12>0

answer: x<-4 and x>-3

according to me it should be x>-3 and x>-4 hence solution should be x>-4 .................

its confusing :? .............please help!!!!!!


Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html
xy-plane-71492.html

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html
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Re: x^2+8x+7<0...................basic stuff but intriguining. [#permalink] New post 25 Nov 2012, 10:11
Thanks Vipps & Bunuel............the concept has become much more simpler.
:lol:
Re: x^2+8x+7<0...................basic stuff but intriguining.   [#permalink] 25 Nov 2012, 10:11
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