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Re: What is root(x^2•y^2) if x < 0 and y > 0? [#permalink]
hey,

Can you please explain why it is negative? I thought when the two number are enclosed within an absolute value bracket, it becomes positive.
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Re: What is root(x^2•y^2) if x < 0 and y > 0? [#permalink]
WOW!!! Thank you for that explanation. I definitely get it now.

:-D :-D :-D :-D :-D :-D :-D 8-)
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What is root(x^2•y^2) if x < 0 and y > 0? [#permalink]
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Hi Nai222,

Since your post was from over a year ago, you might not still be around to read this. This question can be solved by TESTing VALUES:

We're told X < 0 and Y > 0

Let's TEST:
X = -2
Y = +2

So, we'd have \sqrt{(4)(4)} = +4

Using those values in the answers, we get...

Answer A: -(-2)(2) = +4 This IS a match

Answer B: (-2)(2) = -4 NOT a match

Answer C: - |(-2)(2)| = -4 NOT a match

Answer D: |2|(-2) = -4 NOT a match

Answer E: No solution. NOT a match

Final Answer:

GMAT assassins aren't born, they're made,
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Re: What is root(x^2•y^2) if x < 0 and y > 0? [#permalink]
for |xy| we have 4 options

xy, x-y, -xy, -x-y

A
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Re: What is root(x^2•y^2) if x < 0 and y > 0? [#permalink]
We must remember here that x>0 => √x^2 = x else if x is negative => -x
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Re: What is root(x^2•y^2) if x < 0 and y > 0? [#permalink]
Bunuel wrote:
Nai222 wrote:
hey,

Can you please explain why it is negative? I thought when the two number are enclosed within an absolute value bracket, it becomes positive.


\(-xy\) is not negative it's positive. \(xy\) is negative, thus \(-xy=-(negative)=positive\).

THEORY:
\(\sqrt{x^2}=|x|\).

The point here is that as square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).



How is mod of -x equal to -x? Shouldnt it be x?
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What is root(x^2•y^2) if x < 0 and y > 0? [#permalink]
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sinhap07 wrote:
Bunuel wrote:
Nai222 wrote:
hey,

Can you please explain why it is negative? I thought when the two number are enclosed within an absolute value bracket, it becomes positive.

\(-xy\) is not negative it's positive. \(xy\) is negative, thus \(-xy=-(negative)=positive\).

THEORY:
. . .
What function does exactly the same thing [as the square root sign]? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\)...

sinhap07 How is mod of -x equal to -x? Shouldnt it be x?

sinhap07 , I assume you refer to
Quote:
\(|x|= -x\) , if \(x<0\)

For me, the rule can be counterintuitive. I've seen a few ways to think about the rule. Maybe they will help.

If x is negative, then |x| = -x

Possible ways to think about the rule:

1. "The negative of a negative is positive."

Think of the negative sign as signifying "opposite."

That is, the negative sign functions as "the negative of a negative number." And the negative of a negative number is positive. See Bunuel above in bold.

Let x = -3. Per |x| = -x:

|-3| = 3, and +3 is the opposite of -3, thus +3 = -(-3)

2. OR think: "The negative sign on RHS means (-1) multiplied by a negative number \(x\)," thus

|x| = -x
|x| = (-1)(x)
|x| = (-1)(negative #)
|x| = a positive number

3. OR think (similar to #1): "in this rule there is a hidden minus sign."

With a number, the "two negatives" are easy to see

|-3| = 3
|-3| = -(-3)

BUT: |x| = -(x) = -x

With variable \(x\), it is easy to forget that there ARE two negative signs.

With the variable, there is only one minus sign on RHS... because the negative variable \(x\) already "contains" a minus sign.

We just don't (can't) write the minus sign twice with the variable.

|-3| = -(-3) = 3
|x| = -(x) = -x

Those two equations are functionally equivalent.

4. Summary - use any negative number, substituted for x, to see that, if x < 0 , then |x| = -x. Reasons:

|-3| = 3, where +3 is the opposite of -3 [+3 = -(-3)]; RHS is the negative of a negative number

|-3| = 3 = (-1)(-3)

|-3| = -(-3) = 3

The absolute value IS positive (or nonnegative). The sign of a negative variable can obscure that fact.

Hope that helps.
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What is root(x^2•y^2) if x < 0 and y > 0? [#permalink]
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genxer123 wrote:
sinhap07 wrote:
Bunuel wrote:
\(-xy\) is not negative it's positive. \(xy\) is negative, thus \(-xy=-(negative)=positive\).

THEORY:
. . .
What function does exactly the same thing [as the square root sign]? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\)...

sinhap07 How is mod of -x equal to -x? Shouldnt it be x?

sinhap07 , I assume you refer to
Quote:
\(|x|= -x\) , if \(x<0\)

For me, the rule can be counterintuitive. I've seen a few ways to think about the rule. Maybe they will help.

If x is negative, then |x| = -x

Possible ways to think about the rule:

1. "The negative of a negative is positive."

Think of the negative sign as signifying "opposite."

That is, the negative sign functions as "the negative of a negative number." And the negative of a negative number is positive. See Bunuel above in bold.

Let x = -3. Per |x| = -x:

|-3| = 3, and +3 is the opposite of -3, thus +3 = -(-3)

2. OR think: "The negative sign on RHS means (-1) multiplied by a negative number \(x\)," thus

|x| = -x
|x| = (-1)(x)
|x| = (-1)(negative #)
|x| = a positive number

3. OR think (similar to #1): "in this rule there is a hidden minus sign."

With a number, the "two negatives" are easy to see

|-3| = 3
|-3| = -(-3)

BUT: |x| = -(x) = -x

With variable \(x\), it is easy to forget that there ARE two negative signs.

With the variable, there is only one minus sign on RHS... because the negative variable \(x\) already "contains" a minus sign.

We just don't (can't) write the minus sign twice with the variable.

|-3| = -(-3) = 3
|x| = -(x) = -x

Those two equations are functionally equivalent.

4. Summary - use any negative number, substituted for x, to see that, if x < 0 , then |x| = -x. Reasons:

|-3| = 3, where +3 is the opposite of -3; RHS is the negative of a negative number

|-3| = (-1)(-3) = 3

|-3| = -(-3) = 3

The absolute value IS positive (or nonnegative). The sign of a negative variable can obscure that fact.

Hope that helps.


To add:

|-x| = |x|. One way to think about it is that |-x| is the distance between -x and 0 on the number line. Similarly, |x| is the distance between x and 0 on the number line. Obviously -x and x are the same distance from 0. For example, -3 and 3 are the same distance from 0; 2 and -2, are the same distance from 0...

Next, when x is 0 or negative, the rule says that |x| = -x. The absolute value cannot be negative and this rule is not violated here. For example, say x = -10, then |-10| = -(-10) = 10 = positive or generally when x is negative |x| = -x = -negative = positive. Or using the distance concept again |-10| is the distance from -10 to 0, which is 10.

Hope it helps.
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Re: What is root(x^2•y^2) if x < 0 and y > 0? [#permalink]
registerincog wrote:
What is \(\sqrt{x^2 y^2}\) if x < 0 and y > 0?

A) -xy
B) xu
C) -|xy|
D) |y|x
E) No solution



If x<0, then x = (-)
if y > 0, then y = (+)
root x^2y^2 = (+), for, given the exponent is even, the power of (-) is (+)

B) xy = (-)(+) = (-) therefore insufficient
C) (-)(+) = (-)
D)(+)(-) = (-)

A) (-)(-) = (+)

Therefore, the answer is A-xy
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Re: What is root(x^2y^2) if x < 0 and y > 0? [#permalink]
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