krit wrote:
If x^2 + y^2 = 29, what is the value of (x-y)^2
(1) xy = 10
(2) x = 5
Hi guys, I'd like some help here. The answer is A. I see why (1) is sufficient, but I don't get why (2) is not. We can just simply plug x = 5 into the equation, find y, and then solve for (x-y)^2
Thanks
Given: \(x^2+y^2=29\).
Question: \((x-y)^2=x^2-2xy+y^2=(x^2+y^2)-2xy=29-2xy=?\) So, basically we should find the value of \(xy\)
(1) xy = 10. Directly gives us the value we needed. Sufficient.
(2) x = 5. Now even if we substitute the value of \(x\) in \(x^2+y^2=29\) we'll get two values for \(y\): 2 and -2, hence two values for \((x-y)^2\): 9 and 49. Not sufficient.
Answer: A.
Hope it's clear.