1978 AHSME Problems/Problem 9: Difference between revisions
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== Problem 9== | |||
If <math>x<0</math>, then <math>\left|x-\sqrt{(x-1)^2}\right|</math> equals | |||
<math>\textbf{(A) }1\qquad | |||
\textbf{(B) }1-2x\qquad | |||
\textbf{(C) }-2x-1\qquad | |||
\textbf{(D) }1+2x\qquad | |||
\textbf{(E) }2x-1 </math> | |||
== Solution == | |||
We have <math>\sqrt{x^2} = |x|</math>, so we rewrite the expression as follows. | We have <math>\sqrt{x^2} = |x|</math>, so we rewrite the expression as follows. | ||
<cmath>|x - \sqrt{(x-1)^2}| = |x - |x-1||</cmath> | <cmath>|x - \sqrt{(x-1)^2}| = |x - |x-1||</cmath> | ||
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~JustinLee2017 | ~JustinLee2017 | ||
==See Also== | |||
{{AHSME box|year=1978|num-b=8|num-a=10}} | |||
{{MAA Notice}} | |||
Latest revision as of 11:06, 13 February 2021
Problem 9
If 
, then 
equals
Solution
We have 
, so we rewrite the expression as follows.
![\[|x - \sqrt{(x-1)^2}| = |x - |x-1||\]](http://latex.artofproblemsolving.com/8/e/7/8e7e72c2db6d3f885a013df9600391c5dc7caf25.png)
We know that 
, so 
. Thus, we can rewrite 
as 
. So
![\[|x - |x-1|| = |x - (1-x)| = |2x - 1|\]](http://latex.artofproblemsolving.com/9/9/9/999a9fe862cdaebfcc24afb50a94f7939aaee628.png)
.
Since 
. Thus, we can write this as
![\[|2x - 1| = 1- 2x\]](http://latex.artofproblemsolving.com/1/f/3/1f396534d8475085751482a5528df752e5be0af0.png)
~JustinLee2017
See Also
| 1978 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
