Is (x + y)^2 x^2 + y^2?

Question

Is  (x + y)^2 > x^2 + y^2 ?

(1)  |x| \leq 0

(2)  \sqrt{y^2} \leq 0

Official Solution is    HERE .

pushpitkc · Answer

The expression  (x + y)^2 > x^2 + y^2  =  (x^2 + y^2 + 2xy) > x^2 + y^2  =  2xy > 0

(1)  |x| \leq 0 
x can be zero or take positive or negative values.
If x = -1, the expression is false.
If x = 1, the expression is false.
The equation is true only when x=0. 
At x=0, 2xy is never greater than 0. Sufficient

(2)  \sqrt{y^2} \leq 0  
This is the same as saying  |y| \leq 0 
y can be zero or take positive or negative values.
Again going by the options we went for statement 1, we get y=0 as the only solution.
At y=0, the expression 2xy is never greater than 0. Sufficient  (Option D)

niks18 · Answer

the question stem can be reduced to - (x + y)^2 > x^2 + y^2 -------> x^2 + y^2 + 2xy > x^2 + y^2 solving it, the question stem becomes Is xy>0 Statement 1: implies that either |x| = 0 or |x| <0. |x| cannot be negative. so we Know that |x| = 0 or xy=0. Hence we get a NO for our question stem . Sufficient Statement 2: \sqrt{y^2} \leq 0 implies that either \sqrt{y^2} = 0 or \sqrt{y^2} < 0 Square root is always positive in GMAT. Hence y=0. So xy=0. Hence we get a NO for our question stem . Sufficient Hence Option D

shashankism · Answer

Given : nothing 
DS:  (x + y)^2 > x^2 + y^2

Option 1 :  |x| \leq 0  
|x| always assumes +ve value or 0 . so only possibility is x =0. SUFFICIENT

Option 2:  \sqrt{y^2} \leq 0 
 \sqrt{y^2}  always assumes +ve value or 0. So only possibility is y= 0.  SUFFICIENT

BrentGMATPrepNow · Answer

Target question :    Is (x + y)² > x² + y²?

This is a good candidate for  rephrasing the target question .

Take:  (x + y)² > x² + y² 
Expand and simplify:  x² + 2xy + y² > x² + y² 
Subtract x² and y² from both sides to get:  2xy > 0 
Divide both sides by 2 to get:  xy > 0

REPHRASED target question :    Is xy > 0?

Statement 1 : |x| ≤ 0  
This inequality should seem odd to us, since the absolute value of a number is always GREATER THAN OR EQUAL to zero. 
This inequality is saying the absolute value of some number is LESS THAN OR EQUAL to zero. 
Since the absolute value of a number can  never  be negative, it must us the case that |x| = 0, which means  x = 0 . 
If  x = 0 , then xy = ( 0 )y = 0, which means we can answer the  REPHRASED target question  with certainty. 
 It is NOT the case that xy > 0 
Since we can answer the  REPHRASED target question  with certainty, statement 1 is SUFFICIENT

Statement 2 : √(y²) ≤ 0  
This inequality should also seem odd to us, because the square root notation instructs us to find the POSITIVE square root of a number, which means √(some number) is always GREATER THAN OR EQUAL to zero. 
The given inequality is saying √(some number) is LESS THAN OR EQUAL to zero. 
Since √(some number) can  never  be negative, it must us the case that √(y²) = 0, which means  y = 0 . 
If  y = 0 , then xy = (x)( 0 ) = 0, which means we can answer the  REPHRASED target question  with certainty. 
 It is NOT the case that xy > 0 
Since we can answer the  REPHRASED target question  with certainty, statement 2 is SUFFICIENT

Answer:  D

RELATED VIDEO
 https://www.youtube.com/watch?v=MyG1K3ee69w

harikrish · Answer

Solution:

Statement 1: Modulus of a value is never negative.
So the only possible value of x is x=0.

Statement 2: Square root of a value is never negative.
Again the only possible value of x is x=0.

Therefore the answer is Option D.

Bunuel · Answer

Official Solution:

Is  (x + y)^2 > x^2 + y^2 ? 
 
First, let's rephrase the question: is  x^2 + 2xy + y^2 > x^2 + y^2 ?
 
Further simplifying, we arrive at the question: is  xy > 0 ?
 
(1)  |x| \leq 0 . 
 
The absolute value of a number cannot be negative. It can only be 0 or positive. Thus, from this statement, we can deduce that  x = 0 . Therefore,  xy = 0 . This gives us a NO answer to the question. Sufficient.
 
(2)  \sqrt{y^2} \leq 0 . 
 
This is equivalent to  |y| \leq 0 . We have the same as above:  y = 0 . Therefore,  xy = 0 . Again, we have a NO answer to the question. Sufficient.

Answer: D.

sasidharrs · Answer

Hi,
Why sqrt(y2)≤0 is the same as saying |y|≤0 
I know tht sqrt(9) is +/-3 but sqrt(3)^2 is always positive right?

Bunuel · Answer

MUST KNOW: \sqrt{x^2}=|x| : The point here is that since square root function cannot give negative result then \sqrt{some \ expression}\geq{0} . So \sqrt{x^2}\geq{0} . But what does \sqrt{x^2} equal to? Let's consider following examples: If x=5 --> \sqrt{x^2}=\sqrt{25}=5=x=positive ; If x=-5 --> \sqrt{x^2}=\sqrt{25}=5=-x=positive . So we got that: \sqrt{x^2}=x , if x\geq{0} ; \sqrt{x^2}=-x , if x<0 . What function does exactly the same thing? The absolute value function: |x|=x , if x\geq{0} and |x|=-x , if x<0 . That is why \sqrt{x^2}=|x| . Hope it's clear.

bumpbot · Answer

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