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

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07 Sep 2012, 14:48

1

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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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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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09 Sep 2012, 00:59

1

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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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04 Jul 2013, 03:28

1

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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.
_________________

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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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, 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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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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