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

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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>
 
<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>
  
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== Solution ==
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== Solution 1 ==
 
<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>
 
<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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Since <math>x<0,|x|= \boxed{\textbf{(A)}-x} </math>
  
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<math>|x|= -x \Rightarrow A </math>
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== Solution 2 ==
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To confirm the answer, inputting a negative value into <math>x</math> can help. For ease of computation, if <math>x=-3</math>, <math>\sqrt{\frac{-3}{1-\frac{4}{3}}}=\sqrt{\frac{-3}{\frac{-1}{3}}}=\sqrt{9}=3</math>. As no other option choice fits, <math>\boxed{\textbf{(A)}-x}</math> is the correct solution.
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Note that if <math>x=-1</math> was chosen, C would have fit as well. Make sure to avoid making this mistake.
  
 
== See Also ==
 
== See Also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 09:33, 19 August 2022

Problem

Which of the following 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 1

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

Since $x<0,|x|= \boxed{\textbf{(A)}-x}$

Solution 2

To confirm the answer, inputting a negative value into $x$ can help. For ease of computation, if $x=-3$, $\sqrt{\frac{-3}{1-\frac{4}{3}}}=\sqrt{\frac{-3}{\frac{-1}{3}}}=\sqrt{9}=3$. As no other option choice fits, $\boxed{\textbf{(A)}-x}$ is the correct solution. Note that if $x=-1$ was chosen, C would have fit as well. Make sure to avoid making this mistake.

See Also

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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