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Retired Moderator

Joined: **07 Jun 2014 **

Posts: **4804**

WE:**Business Development (Energy and Utilities)**

If x2 - y2 = 0 and xy 0, which of the following MUST be true
[#permalink]
20 May 2018, 10:49

1

Expert Reply

Question Stats:

If \(x^2 - y^2 = 0\) and \(xy \neq 0\), which of the following MUST be true?

Indicate all such statements.

A. x = y

B. |x| = |y|

C. \(\frac{x^2}{y^2}=1\)

Indicate all such statements.

A. x = y

B. |x| = |y|

C. \(\frac{x^2}{y^2}=1\)

Re: If x2 - y2 = 0 and xy 0, which of the following MUST be true
[#permalink]
23 May 2018, 13:40

1

The answers are B and C.

Let's go through each answer one by one.

A. x=y

At first glance, this answer looks correct for the equation \(x^2 - y^2=0\). However, we have to keep in mind that any number, positive or negative, when squared, is positive. Because the only restriction is that \(xy > 0\), this means that either x or y could be positive while the other is a negative.

For example,

When x and y are both 5, then \(5^2 - 5^2 = 0\). However, if x=-5 and y=5, \((-5)^2 - 5^2 = 0\)

So A is false.

B. |x| = |y|

Using the last example, we can see that this is true. |-5| = |5|

The same goes for any combination of numbers.

B is true.

C. \(\frac{x^2}{y^2}=0\)

Following up from A, we already know that any number, positive or negative squared stays positive. If both numbers when squared, subtracted from each other are zero, then one number over the other will always be 1.

Example.

\(\frac{-5^2}{5^2}\) = 0

and \(\frac{5^2}{5^2}\) = 0

C is true.

Let's go through each answer one by one.

A. x=y

At first glance, this answer looks correct for the equation \(x^2 - y^2=0\). However, we have to keep in mind that any number, positive or negative, when squared, is positive. Because the only restriction is that \(xy > 0\), this means that either x or y could be positive while the other is a negative.

For example,

When x and y are both 5, then \(5^2 - 5^2 = 0\). However, if x=-5 and y=5, \((-5)^2 - 5^2 = 0\)

So A is false.

B. |x| = |y|

Using the last example, we can see that this is true. |-5| = |5|

The same goes for any combination of numbers.

B is true.

C. \(\frac{x^2}{y^2}=0\)

Following up from A, we already know that any number, positive or negative squared stays positive. If both numbers when squared, subtracted from each other are zero, then one number over the other will always be 1.

Example.

\(\frac{-5^2}{5^2}\) = 0

and \(\frac{5^2}{5^2}\) = 0

C is true.

Re: If x2 - y2 = 0 and xy 0, which of the following MUST be true
[#permalink]
26 May 2018, 12:59

Good one!

Retired Moderator

Joined: **07 Jun 2014 **

Posts: **4804**

WE:**Business Development (Energy and Utilities)**

If x2 – y2 = 0 and xy ≠ 0, which of the following must be tr
[#permalink]
23 Jun 2018, 12:38

Expert Reply

If \(x^2\) – \(y^2\) = 0 and \(xy \neq 0\), which of the following must be true?

Indicate all such statements.

A. \(x = y\)

B. \(|x| = |y|\)

C. \(\frac{x^2}{y^2}= 1\)

_________________

Indicate all such statements.

A. \(x = y\)

B. \(|x| = |y|\)

C. \(\frac{x^2}{y^2}= 1\)

_________________

Sandy

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If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Retired Moderator

Joined: **07 Jun 2014 **

Posts: **4804**

WE:**Business Development (Energy and Utilities)**

Re: If x2 – y2 = 0 and xy ≠ 0, which of the following must be tr
[#permalink]
18 Jul 2018, 07:10

Expert Reply

Explanation

Since \(x^2 - y^2 = 0\), add \(y^2\) to both sides to get \(x^2 = y^2\). It might look as though x = y, but this is not necessarily the case. For example, x could be 2 and y could be –2.

Algebraically, taking the square root of both sides of x2 = y2 does not yield x = y, but rather |x| = |y|. Thus, the 1st statement is not necessarily true and the 2nd statement is true. The 3rd statement is also true and can be generated algebraically:

\(x^2 - y^2 = 0\)

\(x^2 = y^2\)

\(\frac{x^2}{y^2}= 1\).

_________________

Sandy

If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Since \(x^2 - y^2 = 0\), add \(y^2\) to both sides to get \(x^2 = y^2\). It might look as though x = y, but this is not necessarily the case. For example, x could be 2 and y could be –2.

Algebraically, taking the square root of both sides of x2 = y2 does not yield x = y, but rather |x| = |y|. Thus, the 1st statement is not necessarily true and the 2nd statement is true. The 3rd statement is also true and can be generated algebraically:

\(x^2 - y^2 = 0\)

\(x^2 = y^2\)

\(\frac{x^2}{y^2}= 1\).

_________________

If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Re: If x2 - y2 = 0 and xy 0, which of the following MUST be true
[#permalink]
23 Jul 2018, 03:12

1

sandy wrote:

If \(x^2 - y^2 = 0\) and \(xy \neq 0\), which of the following MUST be true?

Indicate all such statements.

A. x = y

B. |x| = |y|

C. \(\frac{x^2}{y^2}=1\)

Indicate all such statements.

A. x = y

B. |x| = |y|

C. \(\frac{x^2}{y^2}=1\)

Given

\(x^2 - y^2 = 0\)

x^2 = y^2

Note: x and y could be negative or positive. Squared value is equal , not their individual value.

A. clearly out. x=2 , y = -2. Squared value is equal.

B. Squared value has no difference with absolute value. both show the positive answer.

C. True. both x^2 and y^2 have equal value. at the very beginning we got it.

So, B and c are true in all aspects.

Re: If x2 - y2 = 0 and xy 0, which of the following MUST be true
[#permalink]
04 Jul 2019, 10:49

sandy wrote:

If \(x^2 - y^2 = 0\) and \(xy \neq 0\), which of the following MUST be true?

Indicate all such statements.

A. x = y

B. |x| = |y|

C. \(\frac{x^2}{y^2}=1\)

Indicate all such statements.

A. x = y

B. |x| = |y|

C. \(\frac{x^2}{y^2}=1\)

Why We can't take a different value for x and y.? Like: x=2 and y=3, if we so then B. must not be an answer. Only answer C.

_________________

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Re: If x2 - y2 = 0 and xy 0, which of the following MUST be true
[#permalink]
06 Jul 2019, 14:28

huda wrote:

sandy wrote:

Indicate all such statements.

A. x = y

B. |x| = |y|

C. \(\frac{x^2}{y^2}=1\)

Why We can't take a different value for x and y.? Like: x=2 and y=3, if we so then B. must not be an answer. Only answer C.

If you take different values for x andy, then will their square difference yields 0? Certainly not!!

?

Re: If x2 - y2 = 0 and xy 0, which of the following MUST be true
[#permalink]
06 Jul 2019, 14:33

1

sandy wrote:

Indicate all such statements.

A. x = y

B. |x| = |y|

C. \(\frac{x^2}{y^2}=1\)

Let's think this way...

x^2 - y^2 =0

=> (x+y) (x-y)=0

=> x=y or x = -y

So we can eliminate option A. But their absolute value must be equal. We must take B as an answer.

Again,

x^2 - y^2 =0

=> x^2 = y^2

x^2/y^2=1

Definitely C is also our answer! So B,C.

Re: If x2 - y2 = 0 and xy 0, which of the following MUST be true
[#permalink]
08 May 2020, 04:55

PIneappleBoy2 wrote:

The answers are B and C.

Let's go through each answer one by one.

A. x=y

At first glance, this answer looks correct for the equation \(x^2 - y^2=0\). However, we have to keep in mind that any number, positive or negative, when squared, is positive. Because the only restriction is that \(xy > 0\), this means that either x or y could be positive while the other is a negative.

For example,

When x and y are both 5, then \(5^2 - 5^2 = 0\). However, if x=-5 and y=5, \((-5)^2 - 5^2 = 0\)

So A is false.

B. |x| = |y|

Using the last example, we can see that this is true. |-5| = |5|

The same goes for any combination of numbers.

B is true.

C. \(\frac{x^2}{y^2}=0\)

Following up from A, we already know that any number, positive or negative squared stays positive. If both numbers when squared, subtracted from each other are zero, then one number over the other will always be 1.

Example.

\(\frac{-5^2}{5^2}\) = 0

and \(\frac{5^2}{5^2}\) = 0

C is true.

Let's go through each answer one by one.

A. x=y

At first glance, this answer looks correct for the equation \(x^2 - y^2=0\). However, we have to keep in mind that any number, positive or negative, when squared, is positive. Because the only restriction is that \(xy > 0\), this means that either x or y could be positive while the other is a negative.

For example,

When x and y are both 5, then \(5^2 - 5^2 = 0\). However, if x=-5 and y=5, \((-5)^2 - 5^2 = 0\)

So A is false.

B. |x| = |y|

Using the last example, we can see that this is true. |-5| = |5|

The same goes for any combination of numbers.

B is true.

C. \(\frac{x^2}{y^2}=0\)

Following up from A, we already know that any number, positive or negative squared stays positive. If both numbers when squared, subtracted from each other are zero, then one number over the other will always be 1.

Example.

\(\frac{-5^2}{5^2}\) = 0

and \(\frac{5^2}{5^2}\) = 0

C is true.

You accidentally set C. equal to zero instead of 1. Might wanna fix that just so someone on the fence can follow. Thanks!

gmatclubot

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