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03 Apr 2014, 14:50
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What is the value of x?
(1) \(|x|=-\frac{4}{x}\)
(2) \(x=-|\frac{4}{x}|\)
(1) \(|x|=-\frac{4}{x}\)
(2) \(x=-|\frac{4}{x}|\)
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03 Apr 2014, 14:51
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SOLUTION:
What is the value of \(x\)?
(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: \(-\frac{4}{x}\geq{0}\). This means \(x\leq{0}\), but since \(x\) is in the denominator, it cannot be zero. Hence, \(x < 0\).
Next, if \(x < 0\), then \(|x|=-x\). 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: \(|\frac{4}{x}|=-x\).
Similarly here, the left-hand side of the equation is an absolute value, so it must be non-negative. Thus, the right-hand side, \(-x\), must also be non-negative, implying \(-x\geq{0}\). Therefore, \(x\leq{0}\). However, since \(x\) is in the denominator, it cannot be zero, so \(x < 0\).
Next, if \(x < 0\), then \(|x|=-x\). So, \(|\frac{4}{x}|=-x\) becomes \(\frac{4}{-x}=-x\). Simplifying, we get \(x^2=4\), 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
What is the value of \(x\)?
(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: \(-\frac{4}{x}\geq{0}\). This means \(x\leq{0}\), but since \(x\) is in the denominator, it cannot be zero. Hence, \(x < 0\).
Next, if \(x < 0\), then \(|x|=-x\). 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: \(|\frac{4}{x}|=-x\).
Similarly here, the left-hand side of the equation is an absolute value, so it must be non-negative. Thus, the right-hand side, \(-x\), must also be non-negative, implying \(-x\geq{0}\). Therefore, \(x\leq{0}\). However, since \(x\) is in the denominator, it cannot be zero, so \(x < 0\).
Next, if \(x < 0\), then \(|x|=-x\). So, \(|\frac{4}{x}|=-x\) becomes \(\frac{4}{-x}=-x\). Simplifying, we get \(x^2=4\), 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
General Discussion
04 Apr 2014, 04:15
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Bunuel
What is the value of x?
(1) \(|x|=-\frac{4}{x}\)
(2) \(x=-|\frac{4}{x}|\)
(1) \(|x|=-\frac{4}{x}\)
(2) \(x=-|\frac{4}{x}|\)
Sol: St1
|x|= -4/x
Now if x<0 then |x|=-x
and if x> or =0 then |x|=x
For Case 1 we have -x=-4/x or x^2=4 or x=+/-2. Since we took x<0 therefore x=-2
For Case 2 we have x=-4/x or x^2=-4. We know the square of a number is always greater than or equal to 0
therefore St1 is sufficient
For St 2, we have
x=- |4/x|-------> It can be re-written as x+|4/x|=0
Now |4/x|= -4/x if x <0 and
|4/x|=4/x if x>0 therefore
Let us consider 2 cases and we get
For Case 1,we get x-4/x=0 or (x^2-4)/x=0 or x^2=4 or x=+/-2 or x=-2
Cosnider case 2
x+4/x=0 or (x^2+4)/x=0 or x^2=-4 which is not true. Hence x=-2
Ans is D
