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Difference between revisions of "2006 AMC 10B Problems/Problem 7"

 
(Added problem and solution)
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== Problem ==
 
== Problem ==
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Which of the folowing is equivalent to <math> \sqrt{\frac{x}{1-\frac{x-1}{x}}} </math> when <math> x < 0 </math>
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<math> \mathrm{(A) \ } -x\qquad \mathrm{(B) \ } x\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } \sqrt{\frac{x}{2}}\qquad \mathrm{(E) \ } x\sqrt{-1} </math>
 +
 
== Solution ==
 
== Solution ==
 +
<math> \sqrt{\frac{x}{1-\frac{x-1}{x}}} =  \sqrt{\frac{x}{\frac{x}{x}-\frac{x-1}{x}}} = \sqrt{\frac{x}{\frac{x-(x-1)}{x}}} = \sqrt{\frac{x}{\frac{1}{x}}} = \sqrt{x^2} = |x|</math>
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Since <math>x<0</math>
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 +
<math>|x|= -x \Rightarrow A </math>
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== See Also ==
 
== See Also ==
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]

Revision as of 20:43, 13 July 2006

Problem

Which of the folowing is equivalent to $\sqrt{\frac{x}{1-\frac{x-1}{x}}}$ when $x < 0$

$\mathrm{(A) \ } -x\qquad \mathrm{(B) \ } x\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } \sqrt{\frac{x}{2}}\qquad \mathrm{(E) \ } x\sqrt{-1}$

Solution

$\sqrt{\frac{x}{1-\frac{x-1}{x}}} = \sqrt{\frac{x}{\frac{x}{x}-\frac{x-1}{x}}} = \sqrt{\frac{x}{\frac{x-(x-1)}{x}}} = \sqrt{\frac{x}{\frac{1}{x}}} = \sqrt{x^2} = |x|$

Since $x<0$

$|x|= -x \Rightarrow A$

See Also

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