+0  
 
0
101
5
avatar+502 

 

it's hard, i've been stuck for an hour

 Jun 3, 2021
 #1
avatar+114438 
+3

I expect you are over thinking it.

 

if a+b+c=0 and none are 0 (and all are reak)  then one must be neg, one pos and the other either one.  

So when you multiply them together the answer could be positive or negative

 

a/|a|= plus or minus 1    etc

so maybe you are looking for all the possible combinations of

\(\pm 1\pm1\pm1\pm1\)

 

Or maybe you can discount some of those answers off.    Think about it a bit.

Anyway, I know someone else is also answering this right now.

 Jun 3, 2021
 #3
avatar+114438 
+1

If are positive and one neg then you have   

1+1-1-1=0

If 2 are neg and one is pos you have

-1-1+1+1=0

Melody  Jun 3, 2021
 #4
avatar+502 
+1

yes, that makes sense, thanks!

mworkhard222  Jun 3, 2021
 #2
avatar+2193 
+3

There are 2 cases for this (negative, negative, positive) or (positive, positive, negative). 

For this problem, order doesn't matter. 

 

negative, negative, positive

a and b are negative, and c is positive. 

a/|a| would be -1 since a is negative. 

b/|b| would be -1 since b is negative. 

c/|c| would be 1 since c is positive. 

abc/|abc| would be 1, since negative * negative * positive is positive. 

-1 - 1 + 1 + 1 = 0

 

positive, positive, negative

a and b are positive, and c is negative. 

a/|a| would be 1 since a is positive. 

b/|b| would be 1 since b is positive. 

c/|c| would be -1 since c is negative. 

abc/|abc| would be -1, since positive * positive * negative is negative. 

1 + 1 - 1 - 1 = 0

 

So I think there is only one value, 0. 

 

=^._.^=

 Jun 3, 2021
 #5
avatar+502 
+1

tysm!

mworkhard222  Jun 3, 2021

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