Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1978 AHSME Problems/Problem 9"

(Created page with "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> We know that <math>x < 0</math>, so <math>x-1...")
 
 
Line 1: Line 1:
 +
== 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>
Line 9: Line 21:
  
 
~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 $x<0$, then $\left|x-\sqrt{(x-1)^2}\right|$ equals

$\textbf{(A) }1\qquad \textbf{(B) }1-2x\qquad \textbf{(C) }-2x-1\qquad \textbf{(D) }1+2x\qquad \textbf{(E) }2x-1$


Solution

We have $\sqrt{x^2} = |x|$, so we rewrite the expression as follows. \[|x - \sqrt{(x-1)^2}| = |x - |x-1||\] We know that $x < 0$, so $x-1 < 0$. Thus, we can rewrite $|x-1|$ as $1-x$. So \[|x - |x-1|| = |x - (1-x)| = |2x - 1|\]. Since $x< 0, 2x-1 < 0$. Thus, we can write this as \[|2x - 1| = 1- 2x\] $\boxed{B}$


~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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_9&oldid=146435"