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­Replace x+2 with t 

t^2 = |t|
t^2 - |t| = 0
t = 0      t =+-1
x+2 = 0            x+2 = 1        x+2 = -1

x = -2                  x = -1               x = -3

           Product = -6
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Bunuel
What is the product of all the solutions of (x + 2)^2 = |x + 2|?

A. -6
B. -2
C. 2
D. 6
E. 12
­Let, x+2 = a

then (x + 2)^2 = |x + 2| becomes \(a^2 = |a|\) which is possible for

a = ±1 OR a = 0
i.e x+2 = ±1 OR x+2 = 0

if x+2 = 1 then x = -1
if x+2 = -1 then x = -3
if x+2 = 0 then x = -2

Product of all possible values of x = (-1)*(-3)*(-2) = -6

Answer: Option A


­
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First let's understand this, as LHS = RHS.
Both has to be equal so there are only three possibilities i.e put -1, -2, -3.
And other than these LHS will not be equal to RHS.
So ans would be -1*-2*-3 =-6

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Bunuel
What is the product of all the solutions of (x + 2)^2 = |x + 2|?

A. -6
B. -2
C. 2
D. 6
E. 12
­
Keep in mind:

\((x+2)^2 = |x+2|^2\)

Hence we can write the given equation as:\(|x+2|^2 - |x+2| = 0 \)

\(|x+2| * ( |x+2| - 1) = 0\)

So either |x+2| = 0 which means x = -2
or
|x+2| - 1 = 0 which means x is -1 or -3

Product of all solutions = -1 * -2 * -3 = -6

Answer (A)­

Video on solving absolute values:
https://youtu.be/oqVfKQBcnrs
 
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Bunuel
What is the product of all the solutions of (x + 2)^2 = |x + 2|?

A. -6
B. -2
C. 2
D. 6
E. 12
­Here's an easy and quick approach to doing this problem:

We are asked for the product of the solutions, thus the factors of the right option will be the solutions.

It is easy to work with the options given. Say, at random you start off with 12 - Factors of twelve can be written in pairs as follows:
(1,12),(2,6),(3,4) - and negative pairs of the same. Test them out and you will find that they don't work in the given equation.

Do the same for -6, and you'll have both 3 & -2 working. I managed to do this in a flat 25 seconds this way. I found it faster to substitute and do the calcs in my head than to sit and go through the algebra of it.
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O great Bunuel, the math genius, do you have any similar problems for practice?
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braddouglas10
O great Bunuel, the math genius, do you have any similar problems for practice?

Check absolute value questions here: https://gmatclub.com/forum/search.php?s ... &tag_id=58
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Thank you, great Bunuel, the math genius!
Bunuel
braddouglas10
O great Bunuel, the math genius, do you have any similar problems for practice?

Check absolute value questions here: https://gmatclub.com/forum/search.php?s ... &tag_id=58
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