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

(Created page with '==Problem 7== Which of the following equations does NOT have a solution? <math>\text{(A)}\:(x+7)^2=0</math> <math>\text{(B)}\:|-3x|+5=0</math> <math>\text{(C)}\:\sqrt{-x}-2=0<…')
 
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<math>\text{(E)}\:|-3x|-4=0</math>
 
<math>\text{(E)}\:|-3x|-4=0</math>
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== Solution 1 ==
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<math>|-3x|+5=0</math> has no solution because absolute values only output nonnegative numbers.
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Further:
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<math>(x+7)^2 = 0</math> is true for <math>x = -7</math>
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<math>\sqrt{-x}-2=0</math> is true for <math>x = -4</math>
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<math>\sqrt{x}-8=0</math> is true for <math>x = 64</math>
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<math>|-3x|-4=0</math> is true for <math>x = \frac{4}{3}, -\frac{4}{3}</math>
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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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Section 1:
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This contains A,C,D as they are all squares or square roots.
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From skimming, we can get an answer as maybe C
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Section 2:
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This contains B and E
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From skimming we can get we can get answer as maybe B
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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>
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==Video Solution==
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https://youtu.be/9pG49ACG5k8
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~savannahsolver
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== See Also ==
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{{AMC10 box|year=2011|ab=A|num-b=6|num-a=8}}
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{{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

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