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Re: Is (x + y)^2 > x^2 + y^2? [#permalink]
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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) \(\sqrt{y^2} \leq 0\)[/quote]
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)
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Re: Is (x + y)^2 > x^2 + y^2? [#permalink]
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Bunuel wrote:
Is \((x + y)^2 > x^2 + y^2\)?


(1) \(|x| \leq 0\)

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


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
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Re: Is (x + y)^2 > x^2 + y^2? [#permalink]
Bunuel wrote:
Is \((x + y)^2 > x^2 + y^2\)?


(1) \(|x| \leq 0\)

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


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
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Re: Is (x + y)^2 > x^2 + y^2? [#permalink]
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.
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Re: Is (x + y)^2 > x^2 + y^2? [#permalink]
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?
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Re: Is (x + y)^2 > x^2 + y^2? [#permalink]
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sasidharrs wrote:
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?


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.
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Re: Is (x + y)^2 > x^2 + y^2? [#permalink]
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Re: Is (x + y)^2 > x^2 + y^2? [#permalink]
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