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.

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?

sorry. i dint get you.

the question is asking "is sqrt[(x-3)^2] = 3-x?" how can we assume it and deduce the ans..?

Oh right...v just cannot simply the eqn in the question...
Thanks

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?

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..

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

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

got u. Thanks!!! when x<0, it's always YES!!!

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

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.

thanks a lot tendays2go.

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

anyone?

Unless mentioned the sqroot of a number is to be considered positive. That is the sqroot of the number 4 is 2, the principal sqroot. So the sqroot of (x-3)^2=(x-3)