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

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Bunuel
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Is (x + y)^2 > x^2 + y^2? [#permalink] New post   23 Jun 2017, 03:37
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Is \((x + y)^2 > x^2 + y^2\)?


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

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


Official Solution is HERE.

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Bunuel
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Is (x + y)^2 > x^2 + y^2? [#permalink] New post   17 Jul 2017, 06:29
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Bunuel wrote:
Is \((x + y)^2 > x^2 + y^2\)?


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

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


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.
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BrentGMATPrepNow
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Re: Is (x + y)^2 > x^2 + y^2? [#permalink] New post   Updated on: 21 Jul 2020, 08:58
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Bunuel wrote:
Is \((x + y)^2 > x^2 + y^2\)?


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

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


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:
Show Spoiler
D


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Originally posted by BrentGMATPrepNow on 23 Jun 2017, 06:32.
Last edited by BrentGMATPrepNow on 21 Jul 2020, 08:58, edited 1 time in total.
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Re: Is (x + y)^2 > x^2 + y^2? [#permalink] New post   23 Jun 2017, 04:00
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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] New post   23 Jun 2017, 06:13
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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] New post   23 Jun 2017, 06:18
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] New post   24 Jun 2017, 06:48
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] New post   17 Jul 2017, 20:22
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] New post   17 Jul 2017, 21:17
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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] New post   26 Dec 2023, 08:46
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Re: Is (x + y)^2 > x^2 + y^2? [#permalink]
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