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If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
 03 Jul 2012, 21:27
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If |a+b|=|a-b|, then a*b must be equal to: A. 1 B. -1 C. 0 D. 2 E. -2
Last edited by Bunuel on 04 Jul 2012, 01:35, edited 1 time in total.
Edited the question.
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Re: Absolute number. [#permalink]
03 Jul 2012, 21:40
This I solved by trial and error...with a logic that what RHS has (a - b) and LHS has addition of the same terms. So for this to be true one has to be be zero hence answer will be zero.
Regards,
Tushar
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
04 Jul 2012, 01:37
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
04 Jul 2012, 02:47
Clearly if a or b equal zero, ab = 0so, let b\neq{0}Distance of a from b, equals the distance of a from - bDraw this on a number line, a must equal zero same logic holds for a\neq{0}So either a or b = 0, ab = 0or just solve using our normal absolute value method, two cases: (a+b) = (a-b)b = 0-(a+b) = (a-b)-a-b = a-ba = 0so ab = 0
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
07 Sep 2012, 13:39
Bunuel,
Can you always just square absolute values like you did on this problem?
I am wondering if squaring is a valid operation on absolute values.
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
07 Sep 2012, 14:48
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SergeNew wrote: If |a+b|=|a-b|, then a*b must be equal to:
A. 1
B. -1
C. 0
D. 2
E. -2 If b=0, the equality obviously holds. |a+b|=|a-b| means the distance between a and - b is the same as the distance between a and b. For b\neq0, it means that a is the average of - b and b (or the midpoint between - b and b), so necessarily a=0.Altogether, the product ab must be 0.Answer C
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
08 Sep 2012, 02:54
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honggil wrote: Bunuel,
Can you always just square absolute values like you did on this problem?
I am wondering if squaring is a valid operation on absolute values. Check this post: elimination-of-radials-confused-138409.htmlAs for inequalities: A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).For example: 2<4 --> we can square both sides and write: 2^2<4^2; 0\leq{x}<{y} --> we can square both sides and write: x^2<y^2; But if either of side is negative then raising to even power doesn't always work. For example: 1>-2 if we square we'll get 1>4 which is not right. So if given that x>y then we can not square both sides and write x^2>y^2 if we are not certain that both x and y are non-negative. B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example: -2<-1 --> we can raise both sides to third power and write: -2^3=-8<-1=-1^3 or -5<1 --> -5^2=-125<1=1^3; x<y --> we can raise both sides to third power and write: x^3<y^3. Hope it helps.
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
08 Sep 2012, 03:26
honggil wrote: Bunuel,
Can you always just square absolute values like you did on this problem?
I am wondering if squaring is a valid operation on absolute values. Absolute value is always non-negative, so if you have an equality between two absolute values, either they are both 0 or they are equal to the same positive number. Squared, they still remain equal.
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
09 Sep 2012, 00:59
SergeNew wrote: If |a+b|=|a-b|, then a*b must be equal to:
A. 1
B. -1
C. 0
D. 2
E. -2 I did this problem a little differently. Since both sides have absolute values, I used positives and negatives to get the following four possibilities: a+b = a-b a+b = -(a-b) -(a+b) = a-b -(a+b) = -(a-b) Simplifying each equation gets: b = -b a = -a -a = a -b = b From there, I concluded that a*b must be zero since the only way a = -a is if a = 0 and the only way b = -b is if b = 0. Did I do this problem incorrectly?
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
09 Sep 2012, 06:56
HImba88 wrote: SergeNew wrote: If |a+b|=|a-b|, then a*b must be equal to:
A. 1
B. -1
C. 0
D. 2
E. -2 I did this problem a little differently. Since both sides have absolute values, I used positives and negatives to get the following four possibilities: a+b = a-b a+b = -(a-b) -(a+b) = a-b -(a+b) = -(a-b) Simplifying each equation gets: b = -b a = -a -a = a -b = b From there, I concluded that a*b must be zero since the only way a = -a is if a = 0 and the only way b = -b is if b = 0. Did I do this problem incorrectly? It is absolutely correct. Good job!
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
09 Sep 2012, 14:14
EvaJager wrote: HImba88 wrote: SergeNew wrote: If |a+b|=|a-b|, then a*b must be equal to:
A. 1
B. -1
C. 0
D. 2
E. -2 I did this problem a little differently. Since both sides have absolute values, I used positives and negatives to get the following four possibilities: a+b = a-b a+b = -(a-b) -(a+b) = a-b -(a+b) = -(a-b) Simplifying each equation gets: b = -b a = -a -a = a -b = b From there, I concluded that a*b must be zero since the only way a = -a is if a = 0 and the only way b = -b is if b = 0. Did I do this problem incorrectly? It is absolutely correct. Good job! Thanks Eva. I didn't even think initially to square both sides. That way seems much more efficient than the way I approached the problem. Guess my brain is wired differently
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
09 Sep 2012, 15:33
I found the right answer. But not sure if the procedure is right
|a+b| = |a-b|
|a| + |b| = |a| -|b|
|b| = -|b| this is possible only with 0. So it is possible in only 2 cases. 1 and 0. But if it is 1, a+b neq to a-b. so b=0 and a*b = 0. Please let me know if this is a right approach.
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
09 Sep 2012, 16:03
ravipprasad wrote: I found the right answer. But not sure if the procedure is right
|a+b| = |a-b|
|a| + |b| = |a| -|b|
|b| = -|b| this is possible only with 0. So it is possible in only 2 cases. 1 and 0. But if it is 1, a+b neq to a-b. so b=0 and a*b = 0. Please let me know if this is a right approach. Not sure if you can split the absolute value that way. For example: a = -5 b = 3 |a+b| gets you 2 while |a| + |b| gets you 8. That approach does get you the correct answer though so I may be incorrect
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
09 Sep 2012, 21:30
Then definitely my approach is wrong. But looking at the answer choices we can find that it should be 0. Posted from my mobile device 
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
05 Dec 2012, 02:04
Solution 1: Distance perspective
|a-b| = |a+b| ==> The distance of a and b is equal to the distance of a and -b.
<=======(-b)=======0=======(b)======>
Only 0 is the value that has a distance equal to b and -b.
Solution 2:
|a-b| = |a+b| (square both)
a^2 -2ab + b^2 = a^2 + 2ab + b^2
4ab = 0
ab = 0
Answer: 0
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]
07 Dec 2012, 23:59
Can we plug in nos here? when we plug in random nos we realize that this can be equal only when a*b=0
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Re: If |a+b|=|a-b|, then a*b must be equal to:
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07 Dec 2012, 23:59
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