Inequality

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srn: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Thu Feb 05, 2009 7:35 pm; Thanked: 2 times

by srn » Sun Apr 05, 2009 6:16 am

Is sqrt((x-3)^2) = 3 -x ?
a. x is not equal to 3
b. -x * mod x > 0

Ans - b

Please explain how you came to the conclusion.

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Source: Beat The GMAT — Data Sufficiency | Sun Apr 05, 2009 6:16 am

sacx: Master | Next Rank: 500 Posts; Posts: 139; Joined: Wed Oct 22, 2008 4:36 am; Thanked: 17 times

by sacx » Sun Apr 05, 2009 11:13 am

Stmt I

for x = 0

sqrt((0-3)^2) = 3 -0
3 = 3

for x = 4

sqrt((4-3)^2) = 3 -4
1 does not equal -1

for x = -1

sqrt((-1-3)^2) = 3 - (-1)

4 = 4

So for any value of x > 4, the equation is not True

Insufficient

Stmt II

- x * mod x > 0

for this to be true x = negative, since mod x is always positive

eg: x = -3
-(-3) * mod (-3) = 9 > 0
x = 3

-3 * mod (3) = -9 < 0

Now for any negative value of x, the equation is always True

[spoiler]Sufficient

B[/spoiler]

SACX

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scoobydooby: Legendary Member; Posts: 1035; Joined: Wed Aug 27, 2008 10:56 pmFollowed by: 1 member; Thanked: 104 times

by scoobydooby » Sun Apr 05, 2009 11:25 am

am getting D

is sqrt((x-3)^2) = 3 -x ?

simplying the stimulus,
=>(x-3)^(2*1/2)=3-x?
=>x-3=3-x?
=>2x=6 or x=3?

1) x is not equal to 3
sufficient

2) -x * |x| > 0
since |x| always >0, x must be <0
if x<0 , x is a negative number and cant be equal to 3.
sufficient

hence, D

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codesnooker: Legendary Member; Posts: 543; Joined: Fri Jan 18, 2008 1:01 am; Thanked: 43 times; GMAT Score:580

by codesnooker » Sun Apr 05, 2009 4:27 pm

IMO (E).

There is some serious problem with this question. IMO we cannot directly cancel the powers for LHS.

Since, the answer is given as (B), so let me try first to prove statement II as insufficient.

-x * |x| > 0

It clearly means that x should be negative or x < 0

Now suppose, x = -1

then by the given statement, "sqrt((x-3)^2) = 3 -x"

sqrt((-1-3)^2) = 3+1

(-4)^(2/2) = 3+1

if we directly cancels the power for LHS then it means -4 = 4. Which is wrong.

But IMO, we cannot do so, hence, modified equation will be like this...

=> sqrt((-4)^2) = 4
=> sqrt(16) = 4
on doing the square root for LHS, the answer will be +- 4 (i.e. both positive and negative value of 4).

which is again insufficient to deduce the answer.

Same case for statement I.

Hence, in my opinion, answer should be (E) rather (B).

Did I miss something??? Any help will be gratified.

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Sun Apr 05, 2009 4:37 pm

sqrt((-1-3)^2) = 3+1

(-4)^(2/2) = 3+1

if we directly cancels the power for LHS then it means -4 = 4. Which is wrong.

sqrt[(x-3)^2] = |x-3|

|x-3| will be equal to 3-x only when x is negative or 0 or x=3

From stmt II we know x is negative therefore stm II becomes SUFF

Hope this helps!

Regards,
CR

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Musiq: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Thu Apr 10, 2008 1:42 pm; Location: Chicago; Thanked: 20 times

by Musiq » Mon Apr 06, 2009 9:52 am

Very hard question; though I am in agreement with the OA.

This question tests an often forgotten principle that GMAT test-makers love to visually challenge test-takers on. Please see below for an example:

If x^2 = 25, then X has 2 solutions ( +5 and -5)
But,if X = Sqrt (25), then there is only one solution ( +5). This is because it is assumed that SQRT is always positive; which is the case with Real numbers anyway.

So lets, restate the question " Is SQRT ( X - 3) ^ 2 = 3 - X?
LHS can only be positive root; therefore question can be restated as :

Is X -3 = 3-X?

This is the trap that even 700+ test takers will fall into....because even though we are assuming that (X-3), is a positive square root of (X-3) ^2, there is no way of sayng what X is.

So, X - 3 really could be positive, zero or negative depending on what the value of variable X is.

Therefore, the question really boils down to, Is X > +3?

Statement I:
X could be 2 ----- 1 NO
X could be 4 ----- 1 YES
Therefore Insufficiency

Statement II:
Conclusively proves that X is negative. This answers the question with an ALWAYS NO -------> Sufficiency

Therefore answer = B.

I skipped some mathematical steps because other users have already contributed those steps.

For love, not money.

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2010gmat: Master | Next Rank: 500 Posts; Posts: 371; Joined: Fri Mar 06, 2009 2:48 am; Thanked: 27 times; GMAT Score:740

by 2010gmat » Wed Apr 08, 2009 1:22 am

Musiq wrote:Very hard question; though I am in agreement with the OA.

This question tests an often forgotten principle that GMAT test-makers love to visually challenge test-takers on. Please see below for an example:

If x^2 = 25, then X has 2 solutions ( +5 and -5)
But,if X = Sqrt (25), then there is only one solution ( +5). This is because it is assumed that SQRT is always positive; which is the case with Real numbers anyway.
So lets, restate the question " Is SQRT ( X - 3) ^ 2 = 3 - X?
LHS can only be positive root; therefore question can be restated as :

Is X -3 = 3-X?

This is the trap that even 700+ test takers will fall into....because even though we are assuming that (X-3), is a positive square root of (X-3) ^2, there is no way of sayng what X is.

So, X - 3 really could be positive, zero or negative depending on what the value of variable X is.

Therefore, the question really boils down to, Is X > +3?

Statement I:
X could be 2 ----- 1 NO
X could be 4 ----- 1 YES
Therefore Insufficiency

Statement II:
Conclusively proves that X is negative. This answers the question with an ALWAYS NO -------> Sufficiency

Therefore answer = B.

I skipped some mathematical steps because other users have already contributed those steps.

Correct me if i am wrong.. even for real nos. square root of a no. can be -ve .. sqrt 25 can be -5. For real nos. sqrt of negative nos are not defined...like sqrt(-25) is not defined..which is in domain of complex nos....

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Musiq: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Thu Apr 10, 2008 1:42 pm; Location: Chicago; Thanked: 20 times

by Musiq » Wed Apr 08, 2009 11:14 am

2010gmat wrote:
Musiq wrote:Very hard question; though I am in agreement with the OA.

This question tests an often forgotten principle that GMAT test-makers love to visually challenge test-takers on. Please see below for an example:

If x^2 = 25, then X has 2 solutions ( +5 and -5)
But,if X = Sqrt (25), then there is only one solution ( +5). This is because it is assumed that SQRT is always positive; which is the case with Real numbers anyway.
So lets, restate the question " Is SQRT ( X - 3) ^ 2 = 3 - X?
LHS can only be positive root; therefore question can be restated as :

Is X -3 = 3-X?

This is the trap that even 700+ test takers will fall into....because even though we are assuming that (X-3), is a positive square root of (X-3) ^2, there is no way of sayng what X is.

So, X - 3 really could be positive, zero or negative depending on what the value of variable X is.

Therefore, the question really boils down to, Is X > +3?

Statement I:
X could be 2 ----- 1 NO
X could be 4 ----- 1 YES
Therefore Insufficiency

Statement II:
Conclusively proves that X is negative. This answers the question with an ALWAYS NO -------> Sufficiency

Therefore answer = B.

I skipped some mathematical steps because other users have already contributed those steps.
Correct me if i am wrong.. even for real nos. square root of a no. can be -ve .. sqrt 25 can be -5. For real nos. sqrt of negative nos are not defined...like sqrt(-25) is not defined..which is in domain of complex nos....

Sorry for being ambigous ....but it's hard to explain through typing.

I will try again to articulate my thought:

SQRT OF a Single Order Variable(egs: X, Y, a+ b, 3p) on the GMAT is always POSTIVE
SQRT of X^2 has a postive and a negative solution.

In some ways, this goes back to the order of an equation. The number of solutions = the highest exponent of that variable.

So, SQRT of 4 can only be +2.

But, SQRT of 2^2 CAN BE +2 OR -2.

Even though that sounds completely bonkers, it becomes very important when dealing with variables moreso than with constants.

For love, not money.

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