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

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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]
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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