Hi Bunuel,
If we solve both the equation in 1 & 2 as follow-
1) (x + y)^2 = 9a
x^2 + y^2 + 2xy = 9a
x^2 + y^2 = 9a - 2xy
2) (x – y)^2 = a
x^2 + y^2 - 2xy = a
x^2 + y^2 = a + 2xy
Equating both the equations give us
9a - 2xy = a + 2xy
8a = 4xy
2a = xy
Putting this in (1) & (2)
x^2 + y^2 = 9a - 2xy = 9a - 4a = 5a
x^2 + y^2 = a + 2xy = a + 4a = 5a
This providers the solution to the question, hence should'nt the answer be C?
Please let me know if I miscalculated anything here. Thanks!
Bunuel
3. Is x^2 + y^2 > 4a?(1) (x + y)^2 = 9a
(2) (x – y)^2 = a
(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.
(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.
(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.
Answer: E.