Difference between revisions of "1977 AHSME Problems/Problem 8"
(Created page with "==Solution== Solution by e_power_pi_times_i <math>\dfrac{x}{|x} = 1</math> or <math>-1</math> depending whether <math>x</math> is positive or negative. If <math>a</math>, <ma...") |
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| − | <math>\dfrac{x}{|x} = 1</math> or <math>-1</math> depending whether <math>x</math> is positive or negative. If <math>a</math>, <math>b</math>, and <math>c</math> are positive, then the entire thing amounts to <math>4</math>. If one of the three is negative and the other two positive, the answer is <math>0</math>. If two of the three is negative and one is positive, the answer is <math>0</math>. If all three are negative, the answer is <math>-4</math>. Therefore the set is <math>\boxed{\textbf{(B)}\ \{-4,0,4\}}</math>. | + | <math>\dfrac{x}{|x|} = 1</math> or <math>-1</math> depending whether <math>x</math> is positive or negative. If <math>a</math>, <math>b</math>, and <math>c</math> are positive, then the entire thing amounts to <math>4</math>. If one of the three is negative and the other two positive, the answer is <math>0</math>. If two of the three is negative and one is positive, the answer is <math>0</math>. If all three are negative, the answer is <math>-4</math>. Therefore the set is <math>\boxed{\textbf{(B)}\ \{-4,0,4\}}</math>. |
Revision as of 12:17, 18 November 2016
Solution
Solution by e_power_pi_times_i
or 
depending whether 
is positive or negative. If 
, 
, and 
are positive, then the entire thing amounts to 
. If one of the three is negative and the other two positive, the answer is 
. If two of the three is negative and one is positive, the answer is . If all three are negative, the answer is 
. Therefore the set is 
.
