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Re: M13-02 [#permalink]
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prashd wrote:
Hi,

Can you please explain how we get -6 as one of the root in the second case .


|x +2|=4. Two cases:

When x >= -2 we'd have x + 2 = 4 --> x = 2.

When x < -2 we'd have -(x + 2) = 4 --> x = -6.
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Re: M13-02 [#permalink]
why is I x +2 I = 4 an equation to find the second root?
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Re: M13-02 [#permalink]
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rsamant wrote:
why is I x +2 I = 4 an equation to find the second root?


\(||x +2|-2|=2\) --> denote |x+2| as y, so \(|y-2|=2\) --> y - 2 = 2 or -(y - 2) = 2 --> y = 4 or y = 0. Thus \(|x +2|=4\) or \(|x +2|=0\).
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Re: M13-02 [#permalink]
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Bunuel wrote:
How many roots does the equation \(|| x +2 | - 2 | = 2\) have?
A. 0
B. 1
C. 2
D. 3
E. 4


Anything inside an absolute value can be positive or negative. You have to test both. Since there are two absolute values here, there are four possibilities for a root. We need to test (inner)(outer) absolute values which will be ++, +-, -+, or --.

(++)
x+2-2 = 2
x = 2 one root

(+-)
-(x+2-2) = 2
-x-2+2 = 2
x = -2 second root

(-+)
-(x+2)-2 = 2
-x-2-2 = 2
x = -6 third root

(--)
-(-(x+2)-2) = 2
x+2+2=2
x=-2 this root was already found

Three roots, which are x={-6,-2,2}. Answer is D
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Re: M13-02 [#permalink]
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How many roots does the equation ||x+2|−2|=2 have?
A. 0
B. 1
C. 2
D. 3
E. 4

Answer: 1) |x+2|−2 = 2 => |x+2|=4, we get two values of x i.e., x=2 and x=-6.
2) |x+2|−2 = -2 => |x+2|=0, we get only one value for x i.e., x=-2.
Total, we have 3 values for x. So, option (D) is correct.
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Re: M13-02 [#permalink]
I think this is a high-quality question and I agree with explanation. Its a very easy question if one does not get confused by the double mod signs. A qualitative question indeed.
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Re: M13-02 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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M13-02 [#permalink]
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Bunuel wrote:
How many roots does the equation \(|| x +2 | - 2 | = 2\) have?


A. 0
B. 1
C. 2
D. 3
E. 4

­Hi @Bunel,

Please check the below approach and let me know if its correct.
For ||x+2|-2|, lets first consider the case when x+2>=0; x>=-2

Then |x+2-2|=2
|x|=2

Therefore, two solutions are x=+/-2

Next, when x+2<=0, then x<=-2
Then ||x+2|-2|=2 becomes, |-(x+2)-2|
Which becomes, |-x-4|=2
Therefore, |x+4|=2
So, x=-6 or x=-2

Therefore, only 2 solutions i.e. x=-6 or x=2­

Hence, total 3 solutions: -6,-2 and 2
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