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Is (x-3)^2 = 3 - x ? 1) x is not equal to 3 2) x

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Is (x-3)^2 = 3 - x ? 1) x is not equal to 3 2) x [#permalink]

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Is \(\sqrt{(x-3)^2}\) = 3 - x ?

1) x is not equal to 3
2) x < 0

I don't think I even understand what the question is asking. Can you cancel out the square root and the square??
I would think this condition is ALWAYS true, right?

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Re: difficult DS question [#permalink]

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New post 01 Jul 2009, 19:23
Ans should be B

You cannot cancel out square root and square because they do affect the signs of the variables within and will give different results for different negative or positive values.

for sttmt 1 - consider x = 5

we get 2 = -2 (not valid)

if x = -5

we get 8 = 8, (true)

so 1 is not enough

for sttmt 2:

you will find either using logic or taking random values, that if x is negative, LHS is always = RHS.

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Re: difficult DS question [#permalink]

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New post 01 Jul 2009, 19:27
Agree with B.

1) x is not equal to 3
2) x < 0

Any time you see square roots, square both sides...So the question can be rephrased to
Is \((x-3)^2 = (3-x)^2\)?

Now going by rashimnet's explanation, you can see that B is the answer

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Re: difficult DS question [#permalink]

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New post 01 Jul 2009, 19:37
sdrandom1 wrote:
Agree with B.

1) x is not equal to 3
2) x < 0

Any time you see square roots, square both sides...So the question can be rephrased to
Is \((x-3)^2 = (3-x)^2\)?

Now going by rashimnet's explanation, you can see that B is the answer


\((x-3)^2 = (3-x)^2\)

This is exactly what I don't understand. The above is always true!

Say x = 5
LHS: (5-3)^2 = 2^2 = 4
RHS: (3-5)^2 = (-2)^2 = 4

Say x = -5
LHS: (-5-3)^2 = (-8)^2 = 64
RHS: (3+5)^2 = 8^2 = 64

right?

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Re: difficult DS question [#permalink]

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New post 01 Jul 2009, 20:01
I see what u are saying...now even I'm confused.
Reading rashimnet's explanation I understand why B is right.
But not sure if I would have thought the same way, if I had not read the explanation...?

Is there a standard way as to when we should square both and when we shouldn't?

Manhattan book says square both sides whenever you see square roots...so I blindly followed it....and I fell into the trap..

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Re: difficult DS question [#permalink]

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New post 02 Jul 2009, 01:27
thinkblue wrote:
Is \(\sqrt{(x-3)^2}\) = 3 - x ?

1) x is not equal to 3
2) x < 0

I don't think I even understand what the question is asking. Can you cancel out the square root and the square??
I would think this condition is ALWAYS true, right?


yeah..i got is x^2-6x+9=X^2-6x+9 ? which should always be true

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Re: difficult DS question [#permalink]

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New post 02 Jul 2009, 02:58
bigtreezl wrote:
thinkblue wrote:
Is \(\sqrt{(x-3)^2}\) = 3 - x ?

1) x is not equal to 3
2) x < 0

I don't think I even understand what the question is asking. Can you cancel out the square root and the square??
I would think this condition is ALWAYS true, right?


yeah..i got is x^2-6x+9=X^2-6x+9 ? which should always be true


You are somehow wrong. It is not always true that two numbers are equal when their squares are equal.

Look at this example :
-2 is not equal to 2
But -2^2 is equal to 2^2
Same is true for the question.
(x-3)^2 = (3-x)^2 but x-3 is not equal to 3-x ok?

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Re: difficult DS question [#permalink]

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New post 02 Jul 2009, 06:52
thinkblue wrote:
Is \(\sqrt{(x-3)^2}\) = 3 - x ?

1) x is not equal to 3
2) x < 0

I don't think I even understand what the question is asking. Can you cancel out the square root and the square??
I would think this condition is ALWAYS true, right?


S1: X=0, 3=3

X=4, 1=/-1 Insuff.


S2: x<0. Essentially we get sqrt((-x-3)^2)=x+3 taking into account that x is negative.

Thus S2 is true.

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Re: difficult DS question [#permalink]

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New post 02 Jul 2009, 20:47
thinkblue wrote:
Is \(\sqrt{(x-3)^2}\) = 3 - x ?

1) x is not equal to 3
2) x < 0

I don't think I even understand what the question is asking. Can you cancel out the square root and the square??
I would think this condition is ALWAYS true, right?



I'm guessing the answer is E

sqrt((x-3)^2) can either be (x-3) or (3-x)

neither statements tell us whether x-3 can be the answer or 3-x can be the answer

One example of information that would have helped is

(1) sqrt((x-3)^2) is positive
(2) x > 3

in this case the answer would be C ...

but in the case of the question the statements 1 & 2 do not provided any useful information that would allow us to evaluate the equality
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Re: difficult DS question [#permalink]

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New post 03 Jul 2009, 22:04
The point to consider in these type of questions is that the square root can never be -ve.

Now going with he above mentioned fact lets look at LHS and RHS.
LHS = always + ve
RHS = can be +ve or -ve depending upon value of X.
So if any stmt. can guarantee that the RHS will be +ve , then it will be suff.

Lets look a stmt 1 now.
X <> 3
Does it guarantee that RHS will be +ve.No it does not.
because RHS is + if X<3 AND -ve if X > 3. ..................Insuff.

Stmt 2.
Does it guarantee tha RHS is always +ve.
Yes it does.
Hence suff.

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Re: difficult DS question [#permalink]

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New post 06 Jul 2009, 22:23
B is sufficient here

Whenever you see an expression of the type sqrt{x^2} as in this case -> sqrt{(x-3)^2} think of the absolute value of the expression or simply put |x|. And whenever you are dealing with absolute value you need to check both the positive and the negative possibilities.

In this case, Statement I does not help us narrow down whether we are dealing with negative or positive values for x. Statement II, however, does tell us that we are dealing with only negative values so we can answer the question conclusively.

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Re: difficult DS question   [#permalink] 06 Jul 2009, 22:23
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