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555-605 Level|   Algebra|      
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guddo
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 24 Jan 2024, 09:45
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79% (01:11) correct 21% (01:47) wrong based on 611 sessions

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If \(x^2 > y^2\), then \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} =\)

A. \(x - y\)

B. \(x + y\)

C. \(\frac{x - y}{x + y}\)

D. \(\frac{x^2 + y^2}{x + y}\)

E. \(\frac{x^2 - \sqrt{2}xy + y^2}{x+y}\)

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2024-01-24_20-44-00.png
2024-01-24_20-44-00.png [ 15.71 KiB | Viewed 10376 times ]
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A
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 24 Jan 2024, 10:08
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guddo
If \(x^2 > y^2\), then \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} =\)

A. \(x - y\)

B. \(x + y\)

C. \(\frac{x - y}{x + y}\)

D. \(\frac{x^2 + y^2}{x + y}\)

E. \(\frac{x^2 - \sqrt{2}xy + y^2}{x+y}\)

Show Spoiler
Attachment:
2024-01-24_20-44-00.png
Show more

\(x^2 > y^2......x^2-y^2>0\)

\(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} \)

\(\frac{\sqrt{(x^2 -y^2)^2}}{x+y} \)

\(\frac{{x^2 - y^2}}{x+y} \)

\(\frac{(x-y)(x+y)}{x+y} = x-y\)

A
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sbyongid
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 14 Feb 2025, 05:06
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how did the square root in the denominator disappear?
chetan2u
guddo
If \(x^2 > y^2\), then \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} =\)

A. \(x - y\)

B. \(x + y\)

C. \(\frac{x - y}{x + y}\)

D. \(\frac{x^2 + y^2}{x + y}\)

E. \(\frac{x^2 - \sqrt{2}xy + y^2}{x+y}\)

Show Spoiler
Attachment:
2024-01-24_20-44-00.png

\(x^2 > y^2......x^2-y^2>0\)

\(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} \)

\(\frac{\sqrt{(x^2 -y^2)^2}}{x+y} \)

\(\frac{{x^2 - y^2}}{x+y} \)

\(\frac{(x-y)(x+y)}{x+y} = x-y\)

A
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 14 Feb 2025, 11:39
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sbyongid
how did the square root in the denominator disappear?
chetan2u
guddo
If \(x^2 > y^2\), then \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} =\)

A. \(x - y\)

B. \(x + y\)

C. \(\frac{x - y}{x + y}\)

D. \(\frac{x^2 + y^2}{x + y}\)

E. \(\frac{x^2 - \sqrt{2}xy + y^2}{x+y}\)

Show Spoiler
Attachment:
2024-01-24_20-44-00.png

\(x^2 > y^2......x^2-y^2>0\)

\(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} \)

\(\frac{\sqrt{(x^2 -y^2)^2}}{x+y} \)

\(\frac{{x^2 - y^2}}{x+y} \)

\(\frac{(x-y)(x+y)}{x+y} = x-y\)

A
Show more

\(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} \) = \(\frac{Numerator}{Denominator}\)

Numerator = \(\sqrt{x^4 - 2x^2y^2 + y^4}\) Denominator = \(x+y\)

There is no square root in the denominator.
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 17 Feb 2025, 01:33
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guddo
If \(x^2 > y^2\), then \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} =\)

A. \(x - y\)

B. \(x + y\)

C. \(\frac{x - y}{x + y}\)

D. \(\frac{x^2 + y^2}{x + y}\)

E. \(\frac{x^2 - \sqrt{2}xy + y^2}{x+y}\)

Show Spoiler
Attachment:
The attachment 2024-01-24_20-44-00.png is no longer available
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The point in this question is to understand the simple algebric identity which is \((a=b)^2 = a^2 - 2ab + b^2\)

using this identity we can say that \(x^4 - 2x^2y^2 + y^4 = (x^2 -y^2)^2\)

i..e. \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} = \frac{\sqrt{(x^2 - y^2)^2}}{x+y} = \frac{(x^2 - y^2)}{(x+y)} = \frac{(x - y)*(x+y)}{(x+y)}=x-y \)

Answer: Option A
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EnriqueDandolo
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 07 Mar 2025, 13:35
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Hey GMATinsight, what are the steps to go from x^4 - 2x^2y^2 + y^4 to (x^2 - y^2)^2? Personally, I would've had thought erroneously that x^4 - 2^2x^2y^2 +y^4 was equal to (x^2 - y^2)^2.

Many thanks in advance for your kind help and time.

GMATinsight
guddo
If \(x^2 > y^2\), then \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} =\)

A. \(x - y\)

B. \(x + y\)

C. \(\frac{x - y}{x + y}\)

D. \(\frac{x^2 + y^2}{x + y}\)

E. \(\frac{x^2 - \sqrt{2}xy + y^2}{x+y}\)

Show Spoiler
Attachment:
2024-01-24_20-44-00.png

The point in this question is to understand the simple algebric identity which is \((a=b)^2 = a^2 - 2ab + b^2\)

using this identity we can say that \(x^4 - 2x^2y^2 + y^4 = (x^2 -y^2)^2\)

i..e. \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} = \frac{\sqrt{(x^2 - y^2)^2}}{x+y} = \frac{(x^2 - y^2)}{(x+y)} = \frac{(x - y)*(x+y)}{(x+y)}=x-y \)

Answer: Option A
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 08 Mar 2025, 03:52
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EnriqueDandolo
Hey GMATinsight, what are the steps to go from x^4 - 2x^2y^2 + y^4 to (x^2 - y^2)^2? Personally, I would've had thought erroneously that x^4 - 2^2x^2y^2 +y^4 was equal to (x^2 - y^2)^2.

Many thanks in advance for your kind help and time.

GMATinsight
guddo
If \(x^2 > y^2\), then \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} =\)

A. \(x - y\)

B. \(x + y\)

C. \(\frac{x - y}{x + y}\)

D. \(\frac{x^2 + y^2}{x + y}\)

E. \(\frac{x^2 - \sqrt{2}xy + y^2}{x+y}\)

Show Spoiler
Attachment:
2024-01-24_20-44-00.png

The point in this question is to understand the simple algebric identity which is \((a=b)^2 = a^2 - 2ab + b^2\)

using this identity we can say that \(x^4 - 2x^2y^2 + y^4 = (x^2 -y^2)^2\)

i..e. \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} = \frac{\sqrt{(x^2 - y^2)^2}}{x+y} = \frac{(x^2 - y^2)}{(x+y)} = \frac{(x - y)*(x+y)}{(x+y)}=x-y \)

Answer: Option A
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Consider \(x^2 = m\), and \(y^2 = n\). Therefore, \(x^4 = m^2\) and \(y^4 = n^2\)

Substituting these values in \(x^4 - 2x^2y^2 + y^4\) , we get \(m^2 - 2mn + n^2\) which is \((m - n)^2\)

Now, re-substituting the values of m and n, \((m - n)^2\) = \((x^2 -y^2)^2\)
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 09 Mar 2025, 06:49
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Hey Krunaal, just curious, is this how you would have approached the question had you seen it for the first time on the exam? I understand the logic behind it that we recognise the identity x^2-y^2, but at my first go I definitely didn't think of it right away and had to look at other's solutions :/

Krunaal

Consider \(x^2 = m\), and \(y^2 = n\). Therefore, \(x^4 = m^2\) and \(y^4 = n^2\)

Substituting these values in \(x^4 - 2x^2y^2 + y^4\) , we get \(m^2 - 2mn + n^2\) which is \((m - n)^2\)

Now, re-substituting the values of m and n, \((m - n)^2\) = \((x^2 -y^2)^2\)
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 09 Mar 2025, 07:30
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guddo
If \(x^2 > y^2\), then \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} =\)

A. \(x - y\)

B. \(x + y\)

C. \(\frac{x - y}{x + y}\)

D. \(\frac{x^2 + y^2}{x + y}\)

E. \(\frac{x^2 - \sqrt{2}xy + y^2}{x+y}\)

Show Spoiler
Attachment:
2024-01-24_20-44-00.png
Show more

As \(x^2 > y^2\), we can infer that \(x^2 - y^2 > 0\)

\(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y}\)

\(\frac{\sqrt{(x^2-y^2)^2}}{x+y}\)

\(\frac{|x^2-y^2|}{x+y}\)

As \(x^2 - y^2 > 0\), \(|x^2-y^2|\) = \(x^2 - y^2\)

\(\frac{|x^2-y^2|}{x+y}\)

\(\frac{x^2-y^2}{x+y}\)

\(\frac{(x+y)(x-y)}{x+y}\)

= x - y

Option A
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 09 Mar 2025, 07:41
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Can I understand your thought process a bit more? Is x^2 > y^2 a prerequisite for arriving at x^2-y^2? I'm trying to work on being able to see these algebraic identities even when not so obvious.
gmatophobia
guddo
If \(x^2 > y^2\), then \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} =\)

A. \(x - y\)

B. \(x + y\)

C. \(\frac{x - y}{x + y}\)

D. \(\frac{x^2 + y^2}{x + y}\)

E. \(\frac{x^2 - \sqrt{2}xy + y^2}{x+y}\)

Show Spoiler
Attachment:
2024-01-24_20-44-00.png

As \(x^2 > y^2\), we can infer that \(x^2 - y^2 > 0\)

\(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y}\)

\(\frac{\sqrt{(x^2-y^2)^2}}{x+y}\)

\(\frac{|x^2-y^2|}{x+y}\)

As \(x^2 - y^2 > 0\), \(|x^2-y^2|\) = \(x^2 - y^2\)

\(\frac{|x^2-y^2|}{x+y}\)

\(\frac{x^2-y^2}{x+y}\)

\(\frac{(x+y)(x-y)}{x+y}\)

= x - y

Option A
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 09 Mar 2025, 10:33
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BelisariusTirto
Hey Krunaal, just curious, is this how you would have approached the question had you seen it for the first time on the exam? I understand the logic behind it that we recognise the identity x^2-y^2, but at my first go I definitely didn't think of it right away and had to look at other's solutions :/

Krunaal

Consider \(x^2 = m\), and \(y^2 = n\). Therefore, \(x^4 = m^2\) and \(y^4 = n^2\)

Substituting these values in \(x^4 - 2x^2y^2 + y^4\) , we get \(m^2 - 2mn + n^2\) which is \((m - n)^2\)

Now, re-substituting the values of m and n, \((m - n)^2\) = \((x^2 -y^2)^2\)
Show more
Yes, I would approach it this way - a trigger for me would be the square root over numerator, i would want to get rid of it, and an easy way to get rid of it would be if the expression underneath is a square.
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 09 Mar 2025, 14:36
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Hey Krunaal thanks for your help. I was trying to solve the problem the following way, but I am not getting to the answer, could you please help me identify where am I going wrong?

(SQRT(x^4 - 2x^2y^2 + y^4))^2/(x+y)^2

= (x^4 - 2x^2y^2 + y^4)/(x+y)^2

=(x^2 - y^2)^2/(x+y)^2

=(x-y)(x+y)(x-y)(x+y)/(x+y)(x+y)

=(x-y)^2

Thanks in advance for your help.
EnriqueDandolo
Hey GMATinsight, what are the steps to go from x^4 - 2x^2y^2 + y^4 to (x^2 - y^2)^2? Personally, I would've had thought erroneously that x^4 - 2^2x^2y^2 +y^4 was equal to (x^2 - y^2)^2.

Many thanks in advance for your kind help and time.

GMATinsight
guddo
If \(x^2 > y^2\), then \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} =\)

A. \(x - y\)

B. \(x + y\)

C. \(\frac{x - y}{x + y}\)

D. \(\frac{x^2 + y^2}{x + y}\)

E. \(\frac{x^2 - \sqrt{2}xy + y^2}{x+y}\)

Show Spoiler
Attachment:
2024-01-24_20-44-00.png

The point in this question is to understand the simple algebric identity which is \((a=b)^2 = a^2 - 2ab + b^2\)

using this identity we can say that \(x^4 - 2x^2y^2 + y^4 = (x^2 -y^2)^2\)

i..e. \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} = \frac{\sqrt{(x^2 - y^2)^2}}{x+y} = \frac{(x^2 - y^2)}{(x+y)} = \frac{(x - y)*(x+y)}{(x+y)}=x-y \)

Answer: Option A
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 10 Mar 2025, 01:03
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I see, I understand. We want to get rid of the square root and make our life easier. I guess my main point was that from \(x^4 - 2x^2y^2 + y^4\) I wouldn't have inferred we would arrive at \(x^2-y^2 = (x-y)(x+y)\) by factoring \(x^4 - 2x^2y^2 + y^4 --> (x^2-y^2)^2\) to be able to get remove the square root. If anything I would probably have mistaken it for \((x-y)^2 = x^2 + y^2 - 2xy\) as it *seems* more similar ahah :cry:

Would you advise just to keep practicing algebra questions or go back and brush up my theory?
Krunaal
BelisariusTirto
Hey Krunaal, just curious, is this how you would have approached the question had you seen it for the first time on the exam? I understand the logic behind it that we recognise the identity x^2-y^2, but at my first go I definitely didn't think of it right away and had to look at other's solutions :/

Krunaal

Consider \(x^2 = m\), and \(y^2 = n\). Therefore, \(x^4 = m^2\) and \(y^4 = n^2\)

Substituting these values in \(x^4 - 2x^2y^2 + y^4\) , we get \(m^2 - 2mn + n^2\) which is \((m - n)^2\)

Now, re-substituting the values of m and n, \((m - n)^2\) = \((x^2 -y^2)^2\)
Yes, I would approach it this way - a trigger for me would be the square root over numerator, i would want to get rid of it, and an easy way to get rid of it would be if the expression underneath is a square.
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 10 Mar 2025, 06:38
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EnriqueDandolo
Hey Krunaal thanks for your help. I was trying to solve the problem the following way, but I am not getting to the answer, could you please help me identify where am I going wrong?

(SQRT(x^4 - 2x^2y^2 + y^4))^2/(x+y)^2

= (x^4 - 2x^2y^2 + y^4)/(x+y)^2

=(x^2 - y^2)^2/(x+y)^2

=(x-y)(x+y)(x-y)(x+y)/(x+y)(x+y)

=(x-y)^2

Thanks in advance for your help.
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The first line - it will be \(\frac{\sqrt{(x^2 -y^2)^2}}{x+y} \)
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Krunaal
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 10 Mar 2025, 06:55
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BelisariusTirto
I see, I understand. We want to get rid of the square root and make our life easier. I guess my main point was that from \(x^4 - 2x^2y^2 + y^4\) I wouldn't have inferred we would arrive at \(x^2-y^2 = (x-y)(x+y)\) by factoring \(x^4 - 2x^2y^2 + y^4 --> (x^2-y^2)^2\) to be able to get remove the square root. If anything I would probably have mistaken it for \((x-y)^2 = x^2 + y^2 - 2xy\) as it *seems* more similar ahah :cry:

Would you advise just to keep practicing algebra questions or go back and brush up my theory?
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Algebra 101, Math Number Theory - links to some theory you might find helpful. Since you were able to recognise that \(x^4 - 2x^2y^2 + y^4\) is similar to some algebraic identity, you could think a step further and look for what is different between \(x^4 - 2x^2y^2 + y^4\) and \(x^2 - 2xy + y^2\) (the one that you find it similar to) - from there you will realise that x and y are squared, and \((x-y)^2\) will be \((x^2 - y^2)^2 \) in this case.
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EnriqueDandolo
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 10 Mar 2025, 17:08
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Hey Krunaal I understand that it's easier to solve when the first line is as you stated, however I was wondering if it would also be possible to start solving the problem by applying an exponent of 2 to numerator and denominator as I wrote on my past post. Thanks again for your help.

Krunaal
EnriqueDandolo
Hey Krunaal thanks for your help. I was trying to solve the problem the following way, but I am not getting to the answer, could you please help me identify where am I going wrong?

(SQRT(x^4 - 2x^2y^2 + y^4))^2/(x+y)^2

= (x^4 - 2x^2y^2 + y^4)/(x+y)^2

=(x^2 - y^2)^2/(x+y)^2

=(x-y)(x+y)(x-y)(x+y)/(x+y)(x+y)

=(x-y)^2

Thanks in advance for your help.
The first line - it will be \(\frac{\sqrt{(x^2 -y^2)^2}}{x+y} \)
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Krunaal
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 11 Mar 2025, 03:30
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EnriqueDandolo
Hey Krunaal I understand that it's easier to solve when the first line is as you stated, however I was wondering if it would also be possible to start solving the problem by applying an exponent of 2 to numerator and denominator as I wrote on my past post. Thanks again for your help.

Krunaal
EnriqueDandolo
Hey Krunaal thanks for your help. I was trying to solve the problem the following way, but I am not getting to the answer, could you please help me identify where am I going wrong?

(SQRT(x^4 - 2x^2y^2 + y^4))^2/(x+y)^2

= (x^4 - 2x^2y^2 + y^4)/(x+y)^2

=(x^2 - y^2)^2/(x+y)^2

=(x-y)(x+y)(x-y)(x+y)/(x+y)(x+y)

=(x-y)^2

Thanks in advance for your help.
The first line - it will be \(\frac{\sqrt{(x^2 -y^2)^2}}{x+y} \)
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In that case, since you squared at the beginning, you have to take square root at the end to get the final answer

Let \(k\) = \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y}\)

\(k^2\) = \(\frac{x^4 - 2x^2y^2 + y^4}{{(x+y)}^2}\)

Solving, you will get,

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

But, since we want value of k, taking square roots on both sides,

\(k = x-y\)
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sahilsehdev97
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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 18 Sep 2025, 23:23
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Can someone please explain why the answer isn't E? On first glance, the idea is to simplify the exponents, which i did. This seems like a question that has 2 correct answers. Thanks.
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If \(x^2 > y^2\), then \(\frac{\sqrt{x^4 - 2x^2y^2 + y^4}}{x+y} =\)

A. \(x - y\)

B. \(x + y\)

C. \(\frac{x - y}{x + y}\)

D. \(\frac{x^2 + y^2}{x + y}\)

E. \(\frac{x^2 - \sqrt{2}xy + y^2}{x+y}\)

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If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) = 19 Sep 2025, 00:21
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sahilsehdev97
Can someone please explain why the answer isn't E? On first glance, the idea is to simplify the exponents, which i did. This seems like a question that has 2 correct answers. Thanks.

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How are you getting that E is correct?

It’s not E because \(\sqrt{x^4 - 2x^2y^2 + y^4} = \sqrt{(x^2-y^2)^2} = |x^2-y^2|\). Since it’s given that \(x^2 > y^2\), we have \(|x^2-y^2| = x^2-y^2\).

But the denominator in option E, \(x^2 - \sqrt{2}xy + y^2\), does not simplify to \(x^2 - y^2\).
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