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# which of the following must be true? (no restrictions of

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Manager
Joined: 25 Jul 2006
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which of the following must be true? (no restrictions of [#permalink]

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29 May 2007, 17:07
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

which of the following must be true?
(no restrictions of signs of x or y)

I. 0 > -(x+y)^2
II.0 > -(x-y)^2

1. None
2. I only
3. II only
4. I and II
5. Cannot be determined

If you can, please provide an ALGEBRAIC SOLUTION/DERIVATION/PROOF for the two inequations. I got some answers after testing with several numbers - but it took a while and number plug-in method makes me unsure. Thanks.
Manager
Joined: 04 May 2007
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29 May 2007, 20:02
i don't know if I am correct here, where did this Q come from? my order of operations may have been screwed up

my reason: any number squared is positive. the values of x and y are irrelevant to the question. any combination of x and y will either be 1: positive or 2: negative. whatever the value, when it is squared it will be >0. then any positve number multiplied by negative 1 is negative.

then when i thought about it, we don't know if x and y are different numbers. if x and y are the same then 0^2 = 0. so II was out. also, because we don't know the values of x and y, y could =-x then we would have another situation where we are squaring zero. so my final answer is 5 - can not be determined.

please comment on my reasoning... i have been trying to practice with these sorts of problems as they are one of my weaknesses
Manager
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29 May 2007, 20:16
which of the following must be true?
(no restrictions of signs of x or y)

I. 0 > -(x+y)^2
II.0 > -(x-y)^2

1. None
2. I only
3. II only
4. I and II
5. Cannot be determined

I will probably go with 5. Since

For I, If X=-Y then (x+y)=0, the the eqn wont hold true
Similarly for II If x=y, then (x-Y)=0 and eqn wont be true.

I think some information is missing.

Last edited by vijay2001 on 29 May 2007, 20:28, edited 1 time in total.
Manager
Joined: 04 May 2007
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29 May 2007, 20:20
agreed; i think some info might be missing... some additional restrictions would make it more difficult though
Director
Joined: 12 Jun 2006
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29 May 2007, 20:37
Quote:
I. 0 > -(x+y)^2
II.0 > -(x-y)^2

1. None
2. I only
3. II only
4. I and II
5. Cannot be determined

Won't the answer be 1? the right side of the equation is squared and will always be positive (integer, non-integer) unless the right side equals 0. Both instances would result in the above being false.

also, someone mentioned order of operations. won't we distribute the negative signs in both equations before squaring them?
Manager
Joined: 13 Feb 2007
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29 May 2007, 20:54
The answer to that question is none.

i. if x and y are 0, then the inequality doesn't hold.
0 is not greater than 0

ii. if x and y are equal, this doesn't hold either.

i think some info is missing, because the question seems to easy.
Manager
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29 May 2007, 21:01
ggarr,

the order of ops says we should first take care of exponents, and then multiplication. a negative outside parentheses is the same thing as multiplying by -1.

ethan,

i guess the answer is 1 instead of 5; nice catch there. its not that it can't be determined... duh... we determined that the statements aren't true.
Manager
Joined: 04 May 2007
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29 May 2007, 21:01
ggarr,

the order of ops says we should first take care of exponents, and then multiplication. a negative outside parentheses is the same thing as multiplying by -1.

ethan,

i guess the answer is 1 instead of 5; nice catch there. its not that it can't be determined... duh... we determined that the statements aren't true.
Director
Joined: 12 Jun 2006
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29 May 2007, 22:02
Quote:
ggarr,

the order of ops says we should first take care of exponents, and then multiplication. a negative outside parentheses is the same thing as multiplying by -1.
sorry, you're correct. numbers b/w parens is first ... not numbers outside of parens.
29 May 2007, 22:02
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