[/quote]

I expanded option B and equated to A(LHS) and the LHS went 0 and then the whole equation stopped making sense and chooce D.

I am wondering why did you not expanded x+y whole square equation?[/quote]

Let's see what happens if we expand

Given:
QUANTITY A: x - y
QUANTITY B: (x + y)²

Expand Quantity B to get:
QUANTITY A: x - y
QUANTITY B: x² + 2xy + y²

Now what can we do?

Whatever we do from here seems more likely to make matters more confusing.
For example, we COULD add y to both quantities to get:
QUANTITY A: x
QUANTITY B: x² + 2xy + y + y²
This doesn't help.

What if we subtract x from both quantities? We get:
QUANTITY A: 0
QUANTITY B: x² + 2xy + y - x + y²
Yikes!!

At this point, we should probably start considering a different approach.

Cheers,
Brent

The statement say: y < 0 < x

I will pick two numbers and test:

Case 1: x=10, y=-3

Quantity A
x-y =
10-(-3)=
13

Quantity B
(x+y)^2=
(10-3)^2
49

Therefore, B > A

Case 2: x=1/4, y=-1/2

Quantity A
x-y =
1/4 - (-1/2)=
3/4

Quantity B
(x+y)^2=
(1/4-1/2)^2
1/4

Therefore, A > B

So Answer: D

is A> ?
x= 1, y= -1
then Q-A > Q-B

is B>?
x= 10, y= -1
then Q-A < Q-B

QA:
\(x-y=x-(-y)=x+y\) (since y is some negative integer)

QB: \((x+y)^2\)

Recall this rule:
if \(x>1, x^2>x \)
but
if \(0<x<1, x^2<x\)
So if (x+y) is some fraction, then QA is bigger
and if (x+y) is some integer, QB is bigger.

Two cases, therefore D.