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25 Nov 2012, 01:58
Kudos
Bookmarks
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+70
answer: x-3
according to me it should be x>-3 and x>-4 hence solution should be x>-4 .................
its confusing
.............please help!!!!!!
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+70
answer: 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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25 Nov 2012, 02:07
Kudos
Bookmarks
victorywillbemine
Welcome to GMAT club.
.............please help!!!!!!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.
25 Nov 2012, 03:22
Kudos
Bookmarks
My very first reply on the post............yippie 
You cleared the funda completely............. My first kudo to you!!!!!!
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
You cleared the funda completely............. My first kudo to you!!!!!!
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
