If x, y, and z are integers, is x + y^2 + 3z = 0 ?

Question

If x, y, and z are integers, is  x + y^2 + 3z\geq 0  ?

(1)  y\geq{0}

(2)  0\leq{x}  and  0\leq{z}

OE: "Yes. In the given equation, , we know that will always be non-negative We need to know the signs of x and z in order to answer the question."

praveenvino · Answer

In this case, how can we say that value would be greater than zero without knowing the value of y? 
If y =5, x = -20 and z= - 10, then x+y^2+3z would be -20+25-10 which would be a negative value. But if x= -1, z = -2 and y = 5 would give -1+25-6 which would be a positive value.

Bunuel · Answer

I think you are just confused by the notation.

If x, y, and z are integers, is x+y^2+3z >=0 ?

Question: is  x+y^2+3z\geq{0} ? or is the expression  x+y^2+3z  more than or equal to zero. Note that  y^2  is a square of a number hence this term is always more than or equal to zero.

(1)  y\geq{0} . Not sufficient.

(2)  0\leq{x}  and  0\leq{z}  -->  x  and  z  ( 3z ), as well as  y^2 , are more than or equal to zero thus the sum of these three terms must also be more than or equal to zero. Sufficient.

Answer: B.

P.S. Info that x, y, and z are integers is totally irrelevant for this problem.

praveenvino · Answer

Hi Bunuel,

Sorry. It's my bad.

yogirb8801 · Answer

Hi,

Could you please clarify why it is 'B'?

According to 2nd statement. X< or =0 and z < or =0. Therefore, possible values for X={0,-1,-2.....,-10,-13} and same for Z

Given Equation is x+(y^2)+3z

let's say (y^2)=9

Case 1: x=-1,z=-1
(-1)+9+(3*-1)=9+(-4)=5.. i.e >0

Case 2: x=-9,z=-1
(-9)+9+(3*-1)=9+(-12)=-3.. i.e <0

Case 3: x=-3,z=-2
(-3)+9+(3*-2)=9+(-9)=0..

So 2nd statement is insufficient to prove "x+(y^2)+3z >or= 0?"

I can sense that if we put x and z equal to zero then 'x+(y^2)+3z' is greater than zero. According to me, this condition is not satisfied for all x and z values. This question is "must be true" question and not "could be true" question.

Am i missing anything in this?

Thanks-
Yogi

summer101 · Answer

Hey Yogi,
2nd statement says x>=0 and z>=0 not the other way round

yogirb8801 · Answer

oops... ! :S Thanks summer101

Bunuel · Answer

Merging similar topics. Proper question: if-x-y-and-z-are-integers-is-x-y-2-3z-108314.html#p858906 Solution: if-x-y-and-z-are-integers-is-x-y-2-3z-108314.html#p858916 Hope it helps.

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If x, y, and z are integers, is x + y^2 + 3z >= 0 ?

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