If x and y are integers, is 2xy

Question

If x and y are integers, is 2xy < x^2 + y^2?

(1) xy < 0
(2) x + y = 5

oryahalom · Accepted Answer

I found this question nice so i would like to bring my two cents

First lets simplify the inequality: 2xy 0 0<(x-y)^2. So in fact the question asks whether (x-y) are not equal to 0? statement 1: xy<0. This statement tells us that x and y have different signs. Thus (x-y) necessarily couldn't equal to 0 because positive number - negative number = positive number and negative number - positive number = negative number. Sufficient. statement 2: x+y=5. (x-y) could be zero just if both x and y equal to each other. Since x and y are integers we can conclude that x and y are not equal to each other an thus that (x - y) couldn't be equal to 0. Sufficient. Answer: D

nipunjain14 · Answer

st 1: we know (x-y)^2 < 0

thus x^2+y^2 - 2xy<0
given xy<0
then x^2+y^2+2xy<0
x^2+y^2>2xy

Sufficient

St 2:x+y = 5
following can be values of x and y
x = 1, y = 4 or vice versa
for given expression: x^2+y^2 = 1+16 = 17
2xy = 2*1*4 = 8
thus x^2+y^2 > 2xy

another value can be

= 6 y = -1
then x^2+y^2 =36+1 = 37
2xy = 2*-1*6 = -12
thus x^2+y^2 > 2xy
Sufficient

Ans: d

= 6 y = -1 then x^2+y^2 =36+1 = 37 2xy = 2*-1*6 = -12 thus x^2+y^2 > 2xy Sufficient Ans: d

subha107 · Answer

A
Becoz (x-y)^2 >0 is possible only when x not equal to y. In 2 we can not say x not equal to y but in 1 we can say.

Posted from my mobile device

ENGRTOMBA2018 · Answer

in S2, if xy<0 ---> means that x and y have OPPOSITE signs and are NON ZERO. Whatever be the values of x,y, you will always have x  
eq  y

Hope this helps.

dylanl1218 · Answer

Let's rephrase the question to be is X^2 - 2xy + Y^2 > 0, which we can break down even further to be is ABS(X-Y) > 0?.

Before we jump into each statements let's think through the rephrased question a bit more. There's only two possibilities that we need to look out for, either ABS(X-Y) = 0, or ABS(X-Y) > 0, and note that we know that ABS(X-Y) could NEVER be LESS than zero because an abs() is always >=0. So what's the only case in which ABS(X-Y) = 0? Well it's when X=Y, so in other words, we want to find whether or not X=Y.

Statement (1):  
This provides us that x and y have opposite signs and neither is equal to 0. Great so remember that only one number is equal to the opposite, and that's 0, but we know from this statement that neither x or y is 0, so X must NOT be equal to Y. Sufficient.

Statement (2):  
Given the restriction that x and y are integers there is never a situation in which x and y could be equal. Do note that if we were given x+y = 6 then we would have to consider the case that x=3 and y=3, but that's not the case here, so suffecient.

D is the answer.

DrAnkita91 · Answer

2xy<x2+y2
0<x2-2xy+y2
0<(x-y)2
0< (x-y)
y<x

by deriving this I get option E as the answer
Can please anyone clarify, what's wrong with the equation derived?

Bunuel · Answer

All is good up to 0 < (x - y)^2, which is not equivalent to 0 < x - y. The square of a number is always non-negative; hence, 0 < (x - y)^2 is always true unless x = y. When x = y, then (x - y)^2 = 0, not greater than 0. Thus, the question essentially asks whether x = y. If x = y, then the answer to the question is NO. However, if x ≠ y, then the answer to the question is YES.

The first statement, xy < 0, implies that x and y have opposite signs and thus cannot be equal, giving a YES answer to the question and making (1) sufficient.

The second statement, x + y = 5, is also sufficient because we are given that x and y are  integers , and thus for x + y = 5 to be true, they cannot be equal.

Answer: D.

Hope it's clear.

playthegame · Answer

Thanks, for statement two I also ended up writing the inequality as (x+y)^2 minus (x-y)^2 = 4xy so we have x+y = 5.

Or, we have Expression I: 25 minus 4xy = (x-y)^2;

The question is (x-y)^2 > 0? Now, Expression I 25 will always be great than or equal to 4xy as the maximum value of xy given x+y=5 is 25/4 so this statement is sufficent, maximum value occurs when x=y=2.5 which is not possible as x and y are integers.

I was not able to see that the question is asking for x=y so had to go down this route.

Thanks,
Karam

If x and y are integers, is 2xy < x^2 + y^2?

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If x and y are integers, is 2xy < x^2 + y^2?

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