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Re: If a is non-negative, is x^2 + y^2 > 4a ?

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Bunuel wrote:
Marcab wrote:
If a is non-negative, is x^2 + y^2 > 4a ?

(1) (x + y)^2 = 9a
(2) (x - y)^2 = a

Source: Jamboree
I am not convinced with the OA.


If a is non-negative, is x^2 + y^2 > 4a ?

(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.



Hi Bunel,

Can you please explain where am I going wrong:

(x^2+y^2) = x^2+2xy+y^2 = 9a..........(1)

x^2+y^2 >= 2xy ..........(2)

Substitute equation 2 in 1

Thus, (x^2+y^2+x^2+y^2) = 2(x^2+y^2) >= 9a

2(x^2+y^2) >= 9a

Finally, (x^2+y^2) >= 4.5a. Sufficient

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

Answer: A

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