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Bunuel
If x is an integer and \(|||x| - 3| - x| = 3\), what is the sum of all possible values of x ?

A. -6
B. -4
C. -3
D. -1
E. 0

As also mentioned above, the solution for this would be \(x\geq{3}\) and \(0\geq{x}\geq{-3}\)

So sum will be (-3)+(-2)+(-1)+3+4+5....=Infinity
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Bunuel
If x is an integer and \(|||x| - 3| - x| = 3\), what is the sum of all possible values of x ?

A. -6
B. -4
C. -3
D. -1
E. 0

Bunuel There's a bug with this question, all integers greater than 3 can be the solution for this as the left-hand side would simplify to:

\(||x - 3| - x| = |x - 3 - x| = | -3| = 3\).

Added the missing part to the question. Thank you.
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Bunuel
If x is an integer less than or equal to 0 and \(|||x| - 3| - x| = 3\), what is the sum of all possible values of x ?

A. -6
B. -4
C. -3
D. -1
E. 0

As x is an integer less than or equal to 0, |x|=-x
\(|||x| - 3| - x| = 3\)
\(||-x - 3| - x| = 3\)
(I) If -x>3 or x<-3, then -x-3>0, and
\(|-x - 3 - x| = 3\)
Square both sides => \(9+4x^2+12x=9..........4x(x+3)=0\)
So x=0 or x=-3
(II) If -x<3 or x>-3, then -x-3<0, and
\(|3+x - x| = 3\)
So all values of x in this range will fit in => x={0, -1, -2}

Sum of all values = (-3)+(-2)+(-1)+0=-6

A


Hi, can you please explain as to how you made the two cases? I'm quite lost!

Posted from my mobile device
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chetan2u
Bunuel
If x is an integer less than or equal to 0 and \(|||x| - 3| - x| = 3\), what is the sum of all possible values of x ?

A. -6
B. -4
C. -3
D. -1
E. 0

As x is an integer less than or equal to 0, |x|=-x
\(|||x| - 3| - x| = 3\)
\(||-x - 3| - x| = 3\)
(I) If -x>3 or x<-3, then -x-3>0, and
\(|-x - 3 - x| = 3\)
Square both sides => \(9+4x^2+12x=9..........4x(x+3)=0\)
So x=0 or x=-3
(II) If -x<3 or x>-3, then -x-3<0, and
\(|3+x - x| = 3\)
So all values of x in this range will fit in => x={0, -1, -2}

Sum of all values = (-3)+(-2)+(-1)+0=-6

A


Hi, can you please explain as to how you made the two cases? I'm quite lost!

Posted from my mobile device

As x is an integer less than or equal to 0, |x|=-x
\(|||x| - 3| - x| = 3\)
\(||-x - 3| - x| = 3\)

We will have two cases when we open the MODULUS |-x-3|: Either \(-x\geq{3}\) or \(-x<3\)
And the Modulus will behave differently in bot hthe cases.
If the -x-3<0, the mod when opened will have a '-' sign in front.
If -x-3>0, the mod will remain the same when opened.
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chetan2u

(I) If -x>3 or x<-3, then -x-3>0, and
\(|-x - 3 - x| = 3\)
Square both sides => \(9+4x^2+12x=9..........4x(x+3)=0\)
So x=0 or x=-3

Neither 0 nor -3 are less than -3. Why isn't case 1 invalid?
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philipssonicare
chetan2u
If -x>3 or x<-3, then -x-3>0, and
\(|-x - 3 - x| = 3\)
Square both sides => \(9+4x^2+12x=9..........4x(x+3)=0\)
So x=0 or x=-3

Neither 0 nor -3 are less than -3. Why isn't case 1 invalid?


Hi

We have to include -3 too as a value of the x in one of the two cases.
The first case was supposed to be -x is greater than or equal to 3.
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Bunuel
If x is an integer and \(|||x| - 3| - x| = 3\), what is the sum of all possible values of x ?

A. -6
B. -4
C. -3
D. -1
E. 0

As also mentioned above, the solution for this would be \(x\geq{3}\) and \(0\geq{x}\geq{-3}\)

So sum will be (-3)+(-2)+(-1)+3+4+5....=Infinity


It is mentioned in the question that X is integer less tha equal to zero. Hence possible values of X can only be -1,-2 & -3.

The sum would be then -6


Posted from my mobile device
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Rishabhbarmecha
chetan2u
Bunuel
If x is an integer and \(|||x| - 3| - x| = 3\), what is the sum of all possible values of x ?

A. -6
B. -4
C. -3
D. -1
E. 0

As also mentioned above, the solution for this would be \(x\geq{3}\) and \(0\geq{x}\geq{-3}\)

So sum will be (-3)+(-2)+(-1)+3+4+5....=Infinity


It is mentioned in the question that X is integer less tha equal to zero. Hence possible values of X can only be -1,-2 & -3.

The sum would be then -6


Posted from my mobile device

If you read the thread again or the initial question to which this solution was given, you will realise that the question was edited later and the solution for the edited question is -6.
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