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> | ||
− | == Solution == | + | == 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> | ||
− | Since <math>x<0</math> | + | Since <math>x<0,|x|= \boxed{\textbf{(A)}-x} </math> |
− | <math> | + | == Solution 2 == |
+ | 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. | ||
+ | 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 == | ||
− | + | {{AMC10 box|year=2006|ab=B|num-b=6|num-a=8}} | |
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:33, 19 August 2022
Contents
Problem
Which of the following is equivalent to when ?
Solution 1
Since
Solution 2
To confirm the answer, inputting a negative value into can help. For ease of computation, if , . As no other option choice fits, is the correct solution. Note that if 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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