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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]  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: Absolute number. [#permalink]  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]  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: $$(a+b)^2=(a-b)^2$$ --> $$a^2+2ab+b^2=a^2-2ab+b^2$$ --> $$4ab=0$$ --> $$ab=0$$.

Hope it's clear.

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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]  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]  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]  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 $$0.$$

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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]  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]  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]  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]  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]  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]  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]  09 Sep 2012, 15:03
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, 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]  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

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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]  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]  04 Jul 2013, 00:44
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]  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: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]  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)
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Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]  21 Dec 2014, 14:38
I am very noob, please tell me whether this method is wrong or not

|a+b| = |a-b|

Think positive values
so,
a+b = a-b
2b=0 meaning b is 0

or

Think negative, then ,
a+b = - (a-b)
a+b= -a+b
2a=o
a=o

so a*b =0 either way
Re: If |a+b|=|a-b|, then a*b must be equal to:   [#permalink] 21 Dec 2014, 14:38

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