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M30-08

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M30-08  [#permalink]

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New post 16 Sep 2014, 01:45
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

45% (00:53) correct 55% (01:03) wrong based on 29 sessions

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Re M30-08  [#permalink]

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New post 16 Sep 2014, 01:45
Official Solution:


(1) \(|x|=-\frac{4}{x}\).

The left-hand side of the equation is an absolute value, so it must be non-negative, thus the right-hand side must also be non-negative: \(-\frac{4}{x}\geq{0}\), thus \(x\leq{0}\) but since \(x\) is in denominator it cannot be zero, hence \(x < 0\).

Next, \(x < 0\) means that \(|x|=-x\), so \(|x|=-\frac{4}{x}\) becomes \(-x=-\frac{4}{x}\). From this it follows that \(x^2=4\), so \(x=2\) or \(x=-2\). Discard the positive root because we know that \(x\) must be negative and we are left with \(x=-2\). Sufficient.

(2) \(x=-|\frac{4}{x}|\). Re-arrange: \(|\frac{4}{x}|=-x\).

Basically the same here: the left-hand side of the equation is an absolute value, so it must be non-negative, thus the right-hand side must also be non-negative: \(-x\geq{0}\). Re-arrange: \(x\leq{0}\) but since \(x\) is in denominator it cannot be zero, hence \(x < 0\).

Next, \(x < 0\) means that \(|x|=-x\), so \(|\frac{4}{x}|=-x\) becomes \(\frac{4}{-x}=-x\). Simplify: \(x^2=4\), so \(x=2\) or \(x=-2\). Discard the positive root because we know that \(x\) must be negative and we are left with \(x=-2\). Sufficient.


Answer: D
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Re M30-08  [#permalink]

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New post 01 Aug 2016, 03:01
I think this is a high-quality question and I agree with explanation. Great question but must be classified as 700 not 600
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Re: M30-08  [#permalink]

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New post 23 Apr 2018, 04:05
definitely 700 question...
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Re: M30-08 &nbs [#permalink] 23 Apr 2018, 04:05
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