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if x^2-x-6<0, what is the value of x?

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Source: — Data Sufficiency |
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Re: if x^2-x-6<0, what is the value of x?

by logitech » Sat Jan 03, 2009 4:13 pm
ch0719 wrote:if x^2-x-6<0, what is the value of x?

1. x is an positive integer
2. -2<x<2

I found this somewhere on the web, the answer given is A but I thought it should be E, what do you think?
x^2-x-6<0 only when -2 < X < 3

So Statement 1 can not be sufficient. TRY 100 and you wont get a negative number

Statement 2 is a subset of -2 < X < 3 and sufficient.

So I would go with B
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by cramya » Sat Jan 03, 2009 6:20 pm
Let me add to the confusion more :lol:

How abt C)???

x^2-x-6<0
x^2-x<6
x(x-1)<6

or

(x+2) (x-3) < 0

Stmt I

x could be 1,2. No need to analayze any further analysis multiple values x INSUFF

Stmt II

-2<x<2

x could have multiple value satisfying (x+2) (x-3) < 0

-1,0,1.... (just considering integers(not anydecimal values) wihtin this range)

Highly INSUFF

Stmt I and II together

x has to be 1

SUFF

Choose C)
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Re: if x^2-x-6<0, what is the value of x?

by logitech » Sat Jan 03, 2009 6:24 pm
So Cramya you really think that -1,0 wont satisfy the equation?

For 0 it is -6 <0

For -1 it is -4 < 0

So I guess PUFF PUFF PASS :D :D

Oh by the way, what is HIGHLY INSUFF ?
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by cramya » Sat Jan 03, 2009 6:27 pm
Oh by the way, what is HIGHLY INSUFF
->

One that opens up a world of POSSIBILTIES.

Since x did not have to be an integer based on stm II imagine if we had considered all the decimal values between the given range that satisfied the conditions...
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by logitech » Sat Jan 03, 2009 6:31 pm
Cramya,

Did you study Math in the college ? If not, here is the Calculus 101

In fact, if you learn how to construct this tables you will really like them!
Attachments
mstyh.jpg
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by logitech » Sat Jan 03, 2009 6:53 pm
what is the value of x ?

Okay - End of misunderstanding.

Cramya, I thought that you are saying that -1 and 0 will not satisfy the equation. My bad.

And yes the answer is C because we are looking for a VALUE not a RANGE so

any numbers within the range given in B will satisfy the equation but the only POSITIVE integer is + 1, since +2 is not included.
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by ch0719 » Sat Jan 03, 2009 7:23 pm
thanks

given the statement 1 and 2, we can already conclude that the only possible outcome of x is 1, because x has to be a positive int and it has to be lower than a 2, then it got to be 1 regardless of what the question is asking

therefore i guess the question may be not too legit :D
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by cramya » Sat Jan 03, 2009 7:42 pm
therefore i guess the question may be not too legit
What website is the source?
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by vivek.kapoor83 » Sun Jan 04, 2009 1:55 am
cramya wrote:





Cramya
(x+2) (x-3
cramya,
vaulue of X here would be x<-2 and x<3 isnt it but acc to u ..it is
between -2 <x<3 , how cm

if u put x+2<0
u ll get x<-2
or x>-2
Pls explain
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by cramya » Sun Jan 04, 2009 2:00 am
cramya,
vaulue of X here would be x<-2 and x<3 isnt it but acc to u ..it is
between -2 <x<3 , how cm

if u put x+2<0
u ll get x<-2
or x>-2
I dont think I said -2 <x<3 in my post. I am missing something. Let em know.

(x+2) (x-3) < 0 is from the whats given

I was telling -2<x<2 (stmt 2) is insuffcient since -1,0,1 all fit the bill
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by vivek.kapoor83 » Sun Jan 04, 2009 3:06 am
sorry, it was said by Logitech but if logitech has done this, this is not correct...am i rite.
it is basic but i am not confident of my inequality skills
so,
x+2 <0
means x< -2..
am i rite...this is my ques.
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by rajataga » Sun Jan 04, 2009 3:12 am
Hello Vivek,

It is -2<x<3 because, because when you factorize,

(x-3)(x+2)<0

so, only one of the two factors can be negative, and the other one has to positive.

hence, x-3 will be negative only when x<3, and x+2 will be negative, only when x>-2...

hence, we get

-2 < x< 3


even i think that the answer will be C,

since only A and B just narrow down this range, however both combined limit the answer to just 1.
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by tanviet » Sun Jan 04, 2009 6:01 am
x^2 -x-6<0

(x+2)(x-3)<0

-2<x<3

a, not enough

b, not enouth

c, 0<x<2 and x positive interger

x must be 1
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