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 "2011 AMC 10A Problems/Problem 7"
(made sentence more concise)
 
(3 intermediate revisions by 3 users not shown)
Line 16: Line 16:
  
  
−
<math>|-3x|+5=0</math> has no solution because absolute values output positives and this equation implies that the absolute value could output a negative.
+
<math>|-3x|+5=0</math> has no solution because absolute values only output nonnegative numbers.
  
 
Further:
 
Further:
Line 32: Line 32:
 
Instead of solving, we can just categorize and solve.
 
Instead of solving, we can just categorize and solve.
  
−
[hide=Section 1]
+
Section 1:
−
This contains $\mathrm{(A),(C),(D)}
+
This contains A,C,D as they are all squares or square roots.
−
[/hide]
+
From skimming, we can get an answer as maybe C
 +
 
 +
Section 2:
 +
This contains B and E
 +
From skimming we can get we can get answer as maybe B
 +
 
 +
Now we can analyze and we see <math>-x</math> can become <math>x</math> if <math>x=-y</math> and absolute value inequalities cannot be negative, so the answer is <math>\boxed{\mathrm{(B)}}</math>
 +
 
 +
==Video Solution==
 +
https://youtu.be/9pG49ACG5k8
 +
 
 +
~savannahsolver
  
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2011|ab=A|num-b=6|num-a=8}}
 
{{AMC10 box|year=2011|ab=A|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:23, 21 August 2023

Problem 7

Which of the following equations does NOT have a solution?

$\text{(A)}\:(x+7)^2=0$

$\text{(B)}\:|-3x|+5=0$

$\text{(C)}\:\sqrt{-x}-2=0$

$\text{(D)}\:\sqrt{x}-8=0$

$\text{(E)}\:|-3x|-4=0$


Solution 1

$|-3x|+5=0$ has no solution because absolute values only output nonnegative numbers.

Further: $(x+7)^2 = 0$ is true for $x = -7$

$\sqrt{-x}-2=0$ is true for $x = -4$

$\sqrt{x}-8=0$ is true for $x = 64$

$|-3x|-4=0$ is true for $x = \frac{4}{3}, -\frac{4}{3}$

Therefore, the answer is $\boxed{\mathrm{(B)}}$.

Solution 2

Instead of solving, we can just categorize and solve.

Section 1: This contains A,C,D as they are all squares or square roots. From skimming, we can get an answer as maybe C

Section 2: This contains B and E From skimming we can get we can get answer as maybe B

Now we can analyze and we see $-x$ can become $x$ if $x=-y$ and absolute value inequalities cannot be negative, so the answer is $\boxed{\mathrm{(B)}}$

Video Solution

https://youtu.be/9pG49ACG5k8

~savannahsolver

See Also

2011 AMC 10A (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

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=2011_AMC_10A_Problems/Problem_7&oldid=197285"