What is (x^2y^2)^1/2 if x 0 ?

Question

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

KarishmaB · Accepted Answer

Note that  \sqrt{x^2} = |x| . It is not x, it is |x|. When we talk about square root, it implies the principal square root i.e. just the positive square root.

\sqrt{x^2*y^2} = |x|*|y|

Now, if x < 0,  |x| = -x 
If y > 0,  |y| = y

Hence,  |x|*|y| = -x*y

Answer (A)

Bunuel · Answer

-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| .

Bunuel · Answer

If you plug x=-2 and y=2, then  \sqrt{x^2*y^2}=4 :

(A) –xy = 4. Match.
(B) xy = -4. Discard.
(C) –|xy| = -4. Discard.
(D) |y|x = -4. Discard.

So, plunging in gives the same answer.

One could discard B, C and D right away. The square root function cannot give negative result, while all these options are negative, which means that they cannot be correct answers:

(A) –xy = -(negative)(positive) = positive. OK.
(B) xy = (negative)(positive) =  negative . Discard.
(C) –|xy| = -positive =  negative . Discard.
(D) |y|x = positive*negative =  negative . Discard.

Hope it's clear.

Bunuel · Answer

\sqrt{x^2 y^2}=\sqrt{(xy)^2} = |xy| . Since x is negative and y is positive, then xy is negative, thus  |xy|=-xy .

Answer: A.

Nai222 · Answer

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.

Nai222 · Answer

WOW!!! Thank you for that explanation. I definitely get it now. :-D

PareshGmat · Answer

It seems there is a risk if we plugin numbers eg: -2 & 2 which may not give an undisputed answer as computed by Karishma above

Experts, have your say..

KarishmaB · Answer

As explained by Bunuel above, plugging-in numbers isn't necessarily a problem. Sometimes, it can be problematic because multiple options may match. Then you would need to try another set of numbers on the options that matched. Usually, you will get one correct answer within two rounds.

GMatAspirerCA · Answer

Hi Bunuel and Karisma, 
does that mean square root of 4 is always 2 and  not -2 ? 
I thought it had two options just like x to the power 2 = 4 gives, x = 2 and x = -2.

Bunuel · Answer

When the GMAT provides  the square root sign  for an even root, such as  \sqrt{x}  or  \sqrt {x} , then the  only accepted answer is the positive root . That is,  \sqrt{4}=2 , NOT +2 or -2.

In contrast, the equation  x^2=4  has TWO solutions, +2 and -2.

Hope this helps.

EMPOWERgmatRichC · Answer

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:  B

GMAT assassins aren't born, they're made,
Rich

ameyaprabhu · Answer

What's the difference between A and C?

Both will give us the same answer right?

A = - (of a positive xy)
B = - (of a positive xy)

KarishmaB · Answer

You are given that x < 0 (say x is -2) and y > 0 (say y is 5). 
So xy will be -2*5 = -10.

Hence xy will actually be a negative number.

Option (A) = - xy = - (-10) = 10 (a positive number)

Option (C) = - |xy| = - |-10| = -10 (a negative number)

LakerFan24 · Answer

VeritasPrepKarishma

am i correct in saying that  \sqrt{x^{2}}  is NOT the same as writing  \sqrt{(x^{2})} , so is this the reason here that  \sqrt{x^{2}}  = (-x)?

is this what the problem is testing? because otherwise, i cannot understand how an absolute value is negative -- an absolute value can never be negative, because it is the measure of how far away numbers are on a number line

Bunuel · Answer

\sqrt{x^{2}}  is the same as  \sqrt{(x^{2})} .

\sqrt{x^{2}}=|x| . We are given that x is negative. When x < 0, we know that |x| = -x, so  \sqrt{x^{2}}=|x|=-x . Notice that since x is negative, -x = -negative = positive, thus both the square root and the absolute value return positive result.

KarishmaB · Answer

In addition to what Bunuel said above, let me also add a line on how I explain this in words: You are right. An absolute value can never be negative. So |x| will never be negative. But it can be equal to -x. When? When x itself is negative. Note that a variable x CAN stand for a negative value too. So if x itself is negative, -x becomes POSITIVE. And that is the case in which |x| is equal to -x (which is positive).

sinhap07 · Answer

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

generis · Answer

sinhap07  , I assume you refer to
 
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| = 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.

Bunuel · Answer

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.

What is (x^2y^2)^1/2 if x < 0 and y > 0 ?

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