Bunuel
Is \(|x^2 + y^2| > |x^2 - y^2|\)?
(1) \(x > y\)
(2) \(x > 0\)
Since square of any number is ALWAYS non-negative, we can say that:
|x^2 + y^2| > |x^2 - y^2|
will always be true as long as x and y are non-negative
As soon as x = 0 Or y = 0 Or x = y = 0, we have:
|x^2 + y^2| = |x^2 - y^2|
Example:
If y=0: |x^2 + 0^2| = |x^2 - 0^2| = x²
If x=0: |x^2 + y^2| = |x^2 - y^2| = y²
Thus, we need to ensure neither x nor y is zero.
Statement 1: x > y
Not sufficient to say anything since either term can be zero - insufficient
Statement 2: x > 0
Not sufficient to say anything about y since it can be zero - insufficient
Combining:
Not sufficient to say anything since y can still be zero - insufficient
Example:
x = 4, y = 0:
|x^2 + y^2| = |x^2 - y^2| = 16
x = 4, y = 1:
|x^2 + y^2| = 17, that is greater than |x^2 - y^2| = 15
Answer E Posted from my mobile device
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