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

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== Solution ==
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== Solution 1 ==
  
  
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Therefore, the answer is <math> \boxed{\mathrm{(B)}} </math>.
 
Therefore, the answer is <math> \boxed{\mathrm{(B)}} </math>.
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==Solution 2==
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Instead of solving, we can just categorize and solve.
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[hide=Section 1]
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This contains $\mathrm{(A),(C),(D)}
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[/hide]
  
 
== 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}}

Revision as of 21:34, 10 January 2018

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

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

Solution 2

Instead of solving, we can just categorize and solve.

[hide=Section 1] This contains $\mathrm{(A),(C),(D)} [/hide]

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

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