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x is such that x |x| > 0
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19 Dec 2023, 05:51
19 Dec 2023, 05:51
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50% (01:41) wrong based on 20 sessions
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x is such that \(−x |x| > 0\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\( \sqrt{(x-5)^2}\) |
\( [x-5]\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
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Re: x is such that x |x| > 0
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21 Dec 2023, 16:45
21 Dec 2023, 16:45
1
We can interpret that x must be a negative (nonzero) value, since the OPPOSITE of x is greater than 0. (You can ignore the |x| part, since the absolute value will always produce a positive value and doesn't affect the direction of the inequality.)
So, without focusing on the (x-5) inside both Quantity A and B, let's actually consider whether they produce positive or negative numbers. \(\sqrt{(x-5)^2}\) is ALWAYS going to produce a positive number, since the square of every real number will always be positive. (x-5) is ALWAYS going to produce a negative number, since we've interpreted that x is already a negative number.
Therefore, quantity A (a positive number) is greater than quantity B (a negative number).
We can test this by plugging in a negative value for x.
If \(x = -1\),
\(\sqrt{(-1-5)^2} = 6\)
\(-1-5 = -6\)
\(6 > -6\)
So, without focusing on the (x-5) inside both Quantity A and B, let's actually consider whether they produce positive or negative numbers. \(\sqrt{(x-5)^2}\) is ALWAYS going to produce a positive number, since the square of every real number will always be positive. (x-5) is ALWAYS going to produce a negative number, since we've interpreted that x is already a negative number.
Therefore, quantity A (a positive number) is greater than quantity B (a negative number).
We can test this by plugging in a negative value for x.
If \(x = -1\),
\(\sqrt{(-1-5)^2} = 6\)
\(-1-5 = -6\)
\(6 > -6\)
Re: x is such that x |x| > 0
[#permalink]
22 Dec 2023, 02:25
22 Dec 2023, 02:25
1
Since \(−x |x| > 0\) and \(|x|\) will always be positive, \(−x\) also needs to be positive. This means that \(x\) must be negative.
Now if \(x\) is negative, then, \( (x-5)^2\) will always be positive and its square root \( \sqrt{(x-5)^2}\) will also be always positive. We can check this by taking \(x = −1\), we see that \( (x−5)^2\) will be \((−1−5)^2\) which gives us \((−6)^2 = 36\). Now \(\sqrt{(36)} = 6\) which is positive.
If we look at Quantity B, we have \( [x-5]\), and \( [x-5]\) which will always be negative since \(x\) is negative. We can once again check this with \(x = −1\), and we will get \([−1−5]\) which is \(−6\) which is negative.
Quantity A = always positive
Quantity B = always negative
The answer is Quantity A is always greater.
Now if \(x\) is negative, then, \( (x-5)^2\) will always be positive and its square root \( \sqrt{(x-5)^2}\) will also be always positive. We can check this by taking \(x = −1\), we see that \( (x−5)^2\) will be \((−1−5)^2\) which gives us \((−6)^2 = 36\). Now \(\sqrt{(36)} = 6\) which is positive.
If we look at Quantity B, we have \( [x-5]\), and \( [x-5]\) which will always be negative since \(x\) is negative. We can once again check this with \(x = −1\), and we will get \([−1−5]\) which is \(−6\) which is negative.
Quantity A = always positive
Quantity B = always negative
The answer is Quantity A is always greater.
