Is sqrt [(x-3)^2] = 3-x
1) x does not equal to 3
2) -x*lxl > 0
using the first statement..nothing can be found out..so statement one is not sufficient.
before telling u the logic of statement 2 let me tell u how |x| works..
|x| = x when x>0
|x| = -x when x<0..
-x*|x| can have 2 cases:
1) |x| = x:
therefore -x*|x| = -x*x = -(x^2) and hence lessthan 0. which is wrong from statement 2.
2) |x| = -x:
therefore -x*|x| = -x*-x = x^2 which is greater than 0. which is what the statement 2 is telling..
hence in the question sqrt [(x-3)^2] = 3-x
because we know weather x is +tive or -tive..we can answer the question.
hence statement 2 is sufficient.
this is my logic..please correct me if i am wrong.
mod lxl funda?? - GMAT Prep 1
This topic has expert replies
Source: Beat The GMAT — Data Sufficiency |
Thanks got the logic...
But i've got another doubt now
Can v simplify the given eq.?
sqrt[(x-3)^2] = 3-x
=> x-3 = 3-x...
If v simplify this eq., the Ans will be A & NOT B....
So why dont v simplify the given equation?
But i've got another doubt now
Can v simplify the given eq.?
sqrt[(x-3)^2] = 3-x
=> x-3 = 3-x...
If v simplify this eq., the Ans will be A & NOT B....
So why dont v simplify the given equation?
what i don't understand is that the question asks if sqrt [(x-3)^2] = 3-x. How can you answer that by statement 1 or 2?
Statement 2 simply states that x < 0. If statement 2 is sufficient, for all possible values where x < 0, the original statement will either be YES or NO.
Am I not understanding the question correctly?
Statement 2 simply states that x < 0. If statement 2 is sufficient, for all possible values where x < 0, the original statement will either be YES or NO.
Am I not understanding the question correctly?
well according to my logic..
when x< 0 then 3 - (x) will become 3 - (a negative number) which is 3 + (positive number)..so the expression will always be positive.
if x>0 then (3 - x) would have been negative...but a squareroot will never be negative..at least on GMAT.
well i could be wrong..lets hear out from others too..
when x< 0 then 3 - (x) will become 3 - (a negative number) which is 3 + (positive number)..so the expression will always be positive.
if x>0 then (3 - x) would have been negative...but a squareroot will never be negative..at least on GMAT.
well i could be wrong..lets hear out from others too..
Hi kshin78
Lemme attempt 2 explain u this
from statement 2 we get x<0
Now consider the LHS of the enq
LHS = Sqrt[(x-3)^2]
=> Sqrt[(-ve no. - 3)^2]....Consider x to b any -ve no.
=> Sqrt[(-ve no.)^2]
=> Sqrt (+ve no.)
=> + ve no.
Now consider RHS
3-x
=> 3 - (-ve no.)....Substitute x as a -ve no.
=> 3 + any no.
=> + ve no. (Same as the LHS)
Ex. Put x = -4
LHS = Sqrt[(-4-3)^2] = Sqrt[(-7)^2] = Sqrt[49] = 7
RHS = 3-X = 3 - (-4) = 3+4 = 7
Hence proved....Ans is B....satisfied...
Hope this helps...lemme know if still in doubt
Lemme attempt 2 explain u this
from statement 2 we get x<0
Now consider the LHS of the enq
LHS = Sqrt[(x-3)^2]
=> Sqrt[(-ve no. - 3)^2]....Consider x to b any -ve no.
=> Sqrt[(-ve no.)^2]
=> Sqrt (+ve no.)
=> + ve no.
Now consider RHS
3-x
=> 3 - (-ve no.)....Substitute x as a -ve no.
=> 3 + any no.
=> + ve no. (Same as the LHS)
Ex. Put x = -4
LHS = Sqrt[(-4-3)^2] = Sqrt[(-7)^2] = Sqrt[49] = 7
RHS = 3-X = 3 - (-4) = 3+4 = 7
Hence proved....Ans is B....satisfied...
Hope this helps...lemme know if still in doubt
got u. Thanks!!! when x<0, it's always YES!!!rishi235 wrote:Hi kshin78
Lemme attempt 2 explain u this
from statement 2 we get x<0
Now consider the LHS of the enq
LHS = Sqrt[(x-3)^2]
=> Sqrt[(-ve no. - 3)^2]....Consider x to b any -ve no.
=> Sqrt[(-ve no.)^2]
=> Sqrt (+ve no.)
=> + ve no.
Now consider RHS
3-x
=> 3 - (-ve no.)....Substitute x as a -ve no.
=> 3 + any no.
=> + ve no. (Same as the LHS)
Ex. Put x = -4
LHS = Sqrt[(-4-3)^2] = Sqrt[(-7)^2] = Sqrt[49] = 7
RHS = 3-X = 3 - (-4) = 3+4 = 7
Hence proved....Ans is B....satisfied...
Hope this helps...lemme know if still in doubt
-
aspirant_gmat
- Senior | Next Rank: 100 Posts
- Posts: 39
- Joined: Tue Sep 16, 2008 5:55 am
- Location: Hull, UK
- Thanked: 1 times
Hi,
Can anyone explain why we are not taking square root of
(x - 3) ^ 2 as +(x - 3 ) and -(x - 3) ?
Thanks in anticipation,
Rashmi
Can anyone explain why we are not taking square root of
(x - 3) ^ 2 as +(x - 3 ) and -(x - 3) ?
Thanks in anticipation,
Rashmi
-
tendays2go
- Senior | Next Rank: 100 Posts
- Posts: 55
- Joined: Sun Sep 14, 2008 6:51 am
- Location: Netherlands
- Thanked: 10 times
- GMAT Score:680
Kaplan & OG explain this,
sqrt of an integer in GMAT considers only the positive root
so, that's why we are not taking the negative root here.
sqrt of an integer in GMAT considers only the positive root
so, that's why we are not taking the negative root here.
-
aspirant_gmat
- Senior | Next Rank: 100 Posts
- Posts: 39
- Joined: Tue Sep 16, 2008 5:55 am
- Location: Hull, UK
- Thanked: 1 times
-
Fab
- Senior | Next Rank: 100 Posts
- Posts: 87
- Joined: Tue Jul 15, 2008 10:23 am
- Location: Lima
- Thanked: 4 times
- Followed by:1 members
I still have some doubts here, consider this way of solving:
sqrt[(x-3)*(x-3)] = -(x-3)
So, from Statement 2 we can deduct that X is Negative, but:
If we plug numbers (for example X = -3) we will find:
sqrt[(-6)*(-6)] = - (-6)
soLving we have: -6 = 6 .....and this is not true, so statement 2 doesn't really help..
Could someone please be more clear on the explanation..
THANKS
sqrt[(x-3)*(x-3)] = -(x-3)
So, from Statement 2 we can deduct that X is Negative, but:
If we plug numbers (for example X = -3) we will find:
sqrt[(-6)*(-6)] = - (-6)
soLving we have: -6 = 6 .....and this is not true, so statement 2 doesn't really help..
Could someone please be more clear on the explanation..
THANKS
