+0

# Absolute Value

0
101
5
+502

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

Jun 3, 2021

### 5+0 Answers

#1
+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
+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
+502
+1

yes, that makes sense, thanks!

mworkhard222  Jun 3, 2021
#2
+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
+502
+1

tysm!

mworkhard222  Jun 3, 2021