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

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If |a+b|=|a-b|, then a*b must be equal to: [#permalink] New post 03 Jul 2012, 20:27
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Jul 2012, 00:35, edited 1 time in total.
Edited the question.
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink] New post 04 Jul 2012, 00:37
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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

Square both sides: (a+b)^2=(a-b)^2 --> a^2+2ab+b^2=a^2-2ab+b^2 --> 4ab=0 --> ab=0.

Answer: C.

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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink] New post 04 Jul 2012, 01:47
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Clearly if a or b equal zero, ab = 0

so, let b\neq{0}
Distance of a from b, equals the distance of a from -b
Draw this on a number line, a must equal zero

same logic holds for a\neq{0}

So either a or b = 0, ab = 0

or just solve using our normal absolute value method, two cases:

(a+b) = (a-b)
b = 0

-(a+b) = (a-b)
-a-b = a-b
a = 0

so ab = 0
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink] New post 07 Sep 2012, 13:48
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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] New post 08 Sep 2012, 01:54
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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.html

As 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] New post 04 Jul 2013, 02:28
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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.
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Re: Absolute number. [#permalink] New post 03 Jul 2012, 20:40
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.

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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink] New post 07 Sep 2012, 12:39
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] New post 08 Sep 2012, 02:26
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] New post 08 Sep 2012, 23:59
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] New post 09 Sep 2012, 05:56
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] New post 09 Sep 2012, 13:14
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] New post 09 Sep 2012, 14:33
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] New post 09 Sep 2012, 15:03
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] New post 09 Sep 2012, 20:30
Then definitely my approach is wrong. But looking at the answer choices we can find that it should be 0.

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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink] New post 05 Dec 2012, 01:04
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] New post 07 Dec 2012, 22:59
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] New post 04 Jul 2013, 00:44
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink] New post 09 Jul 2013, 16:02
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.

(C)
Re: If |a+b|=|a-b|, then a*b must be equal to:   [#permalink] 09 Jul 2013, 16:02
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