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Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y [#permalink]

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14 Jun 2017, 06:49

1

This post received KUDOS

Bunuel wrote:

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on x-axis

So the question basically asks if Y=0

(x^2 - y^2)^(1/2) = x + y

Squaring both sides

(x+y)(x-y) = (x+y)^2 x-y = x+y y=0

(1) xy is not a square of an integer Since \(-x \leq y \leq x\) and xy is not a square means x is not equal to y But x+y > 0

(\(\sqrt{x^2 - y^2}\) is +ve) => x^2>y^2

If y = 0 then xy=0 which is a square of an integer hence y is not zero and x cannot be 0 because y will be negative then as x>y and then x+y becomes negative but we know that x+y>0

Hence Sufficient

(2) Point (x,y) is not on x axis shows that y can be anything but 0. Sufficient

Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y [#permalink]

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14 Jun 2017, 11:16

Bunuel wrote:

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on x-axis

As per question stem - \(\sqrt{x^2 - y^2} = x + y\). squaring it we get

\(x^2 - y^2= x^2 + y^2 + 2xy\) solving this we get \(y^2+xy=0\) OR y(y+x) = 0 So the question asks us IS y=0 or y=-x

Statement 1: xy is not a square, implies that "y" is not equal to 0 but y can be equal to -x. For e.g x=-2, y=2, then xy = -4, which is not a perfect square as square cannot be negative, or any other value e.g x=2 y=3. Hence insufficient

Statement 2: this implies that x & y are not 0. but x and y can be any number, so y may or may not be equal to -x e.g (2,3) or (2,-2). Hence insufficient

Combining 1 & 2: we know that y is not equal to 0 but "y" and "x" can be any other number. Hence insufficient.

Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y [#permalink]

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15 Jun 2017, 07:42

niks18 wrote:

Bunuel wrote:

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on x-axis

As per question stem - \(\sqrt{x^2 - y^2} = x + y\). squaring it we get

\(x^2 - y^2= x^2 + y^2 + 2xy\) solving this we get \(y^2+xy=0\) OR y(y+x) = 0 So the question asks us IS y=0 or y=-x

Statement 1: xy is not a square, implies that "y" is not equal to 0 but y can be equal to -x. For e.g x=-2, y=2, then xy = -4, which is not a perfect square as square cannot be negative, or any other value e.g x=2 y=3. Hence insufficient

Statement 2: this implies that x & y are not 0. but x and y can be any number, so y may or may not be equal to -x e.g (2,3) or (2,-2). Hence insufficient

Combining 1 & 2: we know that y is not equal to 0 but "y" and "x" can be any other number. Hence insufficient.

Option E

It is mentioned in the stem that -x<y<x. So y cannot be equal to -x.

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on x-axis

As per question stem - \(\sqrt{x^2 - y^2} = x + y\). squaring it we get

\(x^2 - y^2= x^2 + y^2 + 2xy\) solving this we get \(y^2+xy=0\) OR y(y+x) = 0 So the question asks us IS y=0 or y=-x

Statement 1: xy is not a square, implies that "y" is not equal to 0 but y can be equal to -x. For e.g x=-2, y=2, then xy = -4, which is not a perfect square as square cannot be negative, or any other value e.g x=2 y=3. Hence insufficient

Statement 2: this implies that x & y are not 0. but x and y can be any number, so y may or may not be equal to -x e.g (2,3) or (2,-2). Hence insufficient

Combining 1 & 2: we know that y is not equal to 0 but "y" and "x" can be any other number. Hence insufficient.

Option E

It is mentioned in the stem that -x<y<x. So y cannot be equal to -x.

Those questions are awesome. You tagged both question as GMAT Club tests. Have you add them in the tests? i.e should I face them if I take mock exam in GMAT Club tests?

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on x-axis

\(-x \leq y \leq x\) means x is positive or 0.. \(\sqrt{x^2 - y^2} = x + y\) .. Square both sides.. \(x^2-y^2=(x+y)^2=x^2+y^2+2xy.........2y^2+2xy=0......2y(y+x) =0\) So we have to find whether y=0 or x=-y...

I tells us that x is not equal to y.. But can be 0 or x=-y Insuff II tells us y is not equal to 0.. Insuff..

Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y [#permalink]

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15 Jun 2017, 22:09

Bunuel wrote:

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on x-axis

\(\sqrt{x^2 - y^2} = x + y\) - > this result will come about when \(x^2 - y^2\) is either 0 or 1. Let's jump on to the statement analysis. 1) xy is not a square of integer. This means x and y are either different integers or same integers with different signs or they both are 0. Taking all the 3 scenarios, When x=2 and y=-2, \(\sqrt{x^2 - y^2} = x + y\) is true. When x=3 and y=2, \(\sqrt{x^2 - y^2} = x + y\) is not true. When x=0 and y=0, \(\sqrt{x^2 - y^2} = x + y\) is true.

Hence insufficient.

2) Point (x, y) is NOT on x-axis This implies y is not 0, but does not tell anything about x. Hence this statement in itself is insufficient.

Combining Stmt 1 and Stmt 2: Below two scenarios still hold : When x=2 and y=-2, \(\sqrt{x^2 - y^2} = x + y\) is true. When x=3 and y=2, \(\sqrt{x^2 - y^2} = x + y\) is not true.

Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y [#permalink]

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16 Jun 2017, 02:46

chetan2u wrote:

Bunuel wrote:

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer (2) Point (x, y) is NOT on x-axis

\(-x \leq y \leq x\) means x is positive or 0.. \(\sqrt{x^2 - y^2} = x + y\) .. Square both sides.. \(x^2-y^2=(x+y)^2=x^2+y^2+2xy.........2y^2+2xy=0......2y(y+x) =0\) So we have to find whether x=0 or x=-y...

I tells us that x is not equal to y.. But can be 0 or x=-y Insuff II tells us y is not equal to 0.. Insuff..

Combined X can be = -y or not Insufficient

E

I think you mean either y=o or x =-y

Also I have a question if y = 0, is not xy = 0 which is square of an integer zero??

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things: 1. \(x^2-y^2\geq 0\), so the square root of this number will be defined. 2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question: Does \(\sqrt{x^2 - y^2} = x + y\)? Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)? Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\). Does \(y(x+y)=0\)? Does \(y=0\) or \(x=-y\)?

(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0. However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible. Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.

(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0. However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible. Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.

Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y [#permalink]

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23 Jun 2017, 13:34

Bunuel wrote:

OFFICIAL SOLUTION:

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things: 1. \(x^2-y^2\geq 0\), so the square root of this number will be defined. 2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question: Does \(\sqrt{x^2 - y^2} = x + y\)? Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)? Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\). Does \(y(x+y)=0\)? Does \(y=0\) or \(x=-y\)?

(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0. However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible. Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.

(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0. However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible. Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things: 1. \(x^2-y^2\geq 0\), so the square root of this number will be defined. 2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question: Does \(\sqrt{x^2 - y^2} = x + y\)? Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)? Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\). Does \(y(x+y)=0\)? Does \(y=0\) or \(x=-y\)?

(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0. However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible. Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.

(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0. However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible. Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.

Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y [#permalink]

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21 Sep 2017, 01:24

Bunuel wrote:

OFFICIAL SOLUTION:

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things: 1. \(x^2-y^2\geq 0\), so the square root of this number will be defined. 2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question: Does \(\sqrt{x^2 - y^2} = x + y\)? Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)? Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\). Does \(y(x+y)=0\)? Does \(y=0\) or \(x=-y\)?

(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0. However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible. Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.

(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0. However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible. Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.

Answer: E.

I get the answer as B. Can you tell me where I might be making a mistake here: (b) Point (x, y) is NOT on x-axis \(\sqrt{x^2 - y^2} = x + y\) => \(\sqrt{(x-y)(x+y)} = (x+y)\) Dividing both sides by \(\sqrt{(x+y)}\), we get \(\sqrt{(x-y)} = \sqrt{(x+y)}\)

If y = 0 , then x-y = x+y, otherwise they are not equal. Since y != 0(since point (x,y) is not on x axis), x-y not equal to to x+y, hence \(\sqrt{(x-y)} != \sqrt{(x+y)}\) . So, went with B

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things: 1. \(x^2-y^2\geq 0\), so the square root of this number will be defined. 2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question: Does \(\sqrt{x^2 - y^2} = x + y\)? Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)? Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\). Does \(y(x+y)=0\)? Does \(y=0\) or \(x=-y\)?

(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0. However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible. Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.

(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0. However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible. Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.

Answer: E.

I get the answer as B. Can you tell me where I might be making a mistake here: (b) Point (x, y) is NOT on x-axis \(\sqrt{x^2 - y^2} = x + y\) => \(\sqrt{(x-y)(x+y)} = (x+y)\) Dividing both sides by \(\sqrt{(x+y)}\), we get \(\sqrt{(x-y)} = \sqrt{(x+y)}\)

If y = 0 , then x-y = x+y, otherwise they are not equal. Since y != 0(since point (x,y) is not on x axis), x-y not equal to to x+y, hence \(\sqrt{(x-y)} != \sqrt{(x+y)}\) . So, went with B

Please re-read the solution, paying attention to the highlighted part.

As for your solution, you cannot divide by \(\sqrt{(x+y)}\) because it can be 0 for x = -y and we cannot divide by 0.
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