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x2 > y2 and x > –|y|
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02 Sep 2018, 07:32
02 Sep 2018, 07:32
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Question Stats:
70% (01:14) correct
29% (00:52) wrong based on 127 sessions
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\(x^2 > y^2\) and \(x > -|y|\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
x |
y |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: x2 > y2 and x > –|y|
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03 Sep 2018, 10:26
03 Sep 2018, 10:26
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sandy wrote:
\(x^2 > y^2\) and \(x > -|y|\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
x |
y |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Given \(x^2 > y^2\) and \(x > -|y|\)
From \(x^2 > y^2\) => \(x^2\) - \(y^2\) > 0 => ( x + y ) ( x - y ) > 0.
Here ( x + y ) ( x - y ) > 0 either both should be +ve or both -ve to be greater than 0.
But from \(x > -|y|\) we get ( x + y ) > 0 ( Note here |y| is converted to y )
Since we came to know that ( x + y ) > 0 then must be ( x - y ) > 0
So ( x - y ) > 0 => x > y.
A.
General Discussion
Re: x2 > y2 and x > –|y|
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03 Sep 2018, 15:40
03 Sep 2018, 15:40
2
if \(x^2>y^2\), then \(|x|>|y|\)
if x>0 y>0, then x>y since \(|x|>|y|\);
if x>0 y<0, then x>y. Because \(x>-|y|\) and \(-|y|=-(-y)=y\);
x<0 y>0 is not possible. Because \(x>-|y|\) and \(-|y|=-y\), so \(x>-y\), which means \(|x|<|y|\), but we already know \(|x|>|y|\), so this is a contradiction;
x<0 y<0 is also not possbile, Because \(x>-|y|\) and \(-|y|=-(-y)=y\) therefore x>y but then \(|x|<|y|\), and we already know \(|x|>|y|\), so this is a contradiction;
if x>0 y>0, then x>y since \(|x|>|y|\);
if x>0 y<0, then x>y. Because \(x>-|y|\) and \(-|y|=-(-y)=y\);
x<0 y>0 is not possible. Because \(x>-|y|\) and \(-|y|=-y\), so \(x>-y\), which means \(|x|<|y|\), but we already know \(|x|>|y|\), so this is a contradiction;
x<0 y<0 is also not possbile, Because \(x>-|y|\) and \(-|y|=-(-y)=y\) therefore x>y but then \(|x|<|y|\), and we already know \(|x|>|y|\), so this is a contradiction;
