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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.

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

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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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PhD in Applied Mathematics Love GMAT Quant questions and running.

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: \(x^3<y^3\).

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

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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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PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

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09 Sep 2012, 00:59

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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

I did this problem a little differently. Since both sides have absolute values, I used positives and negatives to get the following four possibilities:

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?

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

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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:

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!
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

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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:

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

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

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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.

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

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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

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

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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: [#permalink]

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04 Jul 2013, 03:28

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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

Think the equation without "a". We know that -b and +b are equal in magnitude. We are adding "a" to each. That still doesn't change the equality of the magnitude. That is possible only when 0 is added or "a" is 0. We can say the reverse also and say that to a , we add b and -b and the equality still holds. In this case b is 0. Either a or b is 0. So a*b must be equal to 0.
_________________

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

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09 Jul 2013, 17:02

If |a+b|=|a-b|, then a*b must be equal to:

|a+b|=|a-b| (a+b)*(a+b) = (a-b)*(a-b) a^2+2ab+b^2 = a^2-2ab+b^2 4ab=0 In other words, a or b must = 0, therefore, the product of a*b is 0 regardless of what a or b are...one of them is 0.

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