Difference between revisions of "2011 AMC 10A Problems/Problem 7"

(Problem 7)
m (Problem 7)
Line 12: Line 12:
 
<math>\text{(E)}\:|-3x|-4=0</math>
 
<math>\text{(E)}\:|-3x|-4=0</math>
  
----
+
 
 +
== Solution ==
 +
 
  
 
<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 output positives and this equation implies that the absolute value could output a negative.
Line 24: Line 26:
  
 
<math>|-3x|-4=0</math> is true for <math>x = \frac{4}{3}, -\frac{4}{3}</math>
 
<math>|-3x|-4=0</math> is true for <math>x = \frac{4}{3}, -\frac{4}{3}</math>
 +
== See Also ==
 +
{{AMC10 box|year=2011|ab=A|num-b=6|num-a=8}}

Revision as of 10:54, 8 May 2011

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

$|-3x|+5=0$ has no solution because absolute values output positives and this equation implies that the absolute value could output a negative.

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 $x = \frac{4}{3}, -\frac{4}{3}$

See Also

2011 AMC 10A (ProblemsAnswer KeyResources)
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