If x^2 y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) =

Question

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} 2024-01-24_20-44-00.png

chetan2u · Accepted Answer

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

sbyongid · Answer

how did the square root in the denominator disappear?

Krunaal · Answer

\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.

GMATinsight · Answer

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

EnriqueDandolo · Answer

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.

Krunaal · Answer

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

BelisariusTirto · Answer

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 :/

gmatophobia · Answer

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

BelisariusTirto · Answer

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.

Krunaal · Answer

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.

EnriqueDandolo · Answer

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.

BelisariusTirto · Answer

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

Would you advise just to keep practicing algebra questions or go back and brush up my theory?

Would you advise just to keep practicing algebra questions or go back and brush up my theory?

Krunaal · Answer

The first line - it will be  \frac{\sqrt{(x^2 -y^2)^2}}{x+y}

Krunaal · Answer

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.

EnriqueDandolo · Answer

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 · Answer

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

sahilsehdev97 · Answer

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.

Bunuel · Answer

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 .

repellaterror · Answer

I recommend using smart numbers here. If you assume x=3 then x^2 is 9, and if you assume y=2 then y^2=4. When you plug and chug, you solve and get the value of 1. x-y=1 since x-y=3-2. Hope this helps, way easier to just plug in this situations than do out all the algebra.

If x^2 > y^2, then (x^4 - 2x^2y^2 + y^4)^(1/2)/(x + y) =

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