Bunuel wrote:
Is x^2 + y^2 > 3z
(1) (x + y)^2 = 9z and (x - y)^2 = z --> \(x^2+2xy+y^2=9z\) and \(x^2-2xy+y^2=z\). Add them up \(2(x^2+y^2)=10z\) --> \(x^2+y^2=5z\). If \(x=y=z=0\), then \(x^2 + y^2=0=3z\) and the answer is NO but if x, y, and z are different from zero, then the answer is YES. Not sufficient.
(2) z = 0. If \(x=y=z=0\), then \(x^2 + y^2=0=3z\) and the answer is NO but if x or y are different from zero, then the answer is YES. Not sufficient.
(1)+(2) From (1) we have that \(x^2+y^2=(non \ negative)+(non \ negative)=5z\) and since from (2) we have that \(z=0\), then \(x=y=z=0\). Thereofre the answer to the question is NO. Sufficient.
Answer: C.
Hi
Bunuel, is my explanation ok? Thank you brother...
statement 1:
x^2+y^2+2xy=9z
x^2+y^2-2xy=z
if we add them then we will get the following
2x^2+2y^2=10z
---> x^2+y^2=5z
Now the question is :
IS x^2+y^2>3z?
So, the rephrase question is-----
IS 5z>3z?
if z=1, then 5*1>3*1?
5>3?
yes, 5 is always greater than 3,
again,
5z>3z?
if we put the value of z=-1, then we get...
5*(-1)>3*(-1)?
-5>-3?
No, -5 is always smaller than -3
So, statement 1 is not sufficient...
Statement 2:
if you put the value of Z=0, then the question stem will be.....
x^2+y^2>3*0?
---> x^2+y^2>0?
if we put positive value for x and y, it always give positive result and also if we put negative value for x and y, it also give positive value as a nature of POSITIVE power. But, if we let x=0, and y=0, then the equation will be----
0^2+0^2>0?
0+0>0?
0 >0?
No,
---> not sufficient.
(1)+(2)
5z>3z?
5*0>3*0?
0>0?
No,
---->sufficient.
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