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Bunuel
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\(­|a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} =-|c| \)­

As we are adding an absolute value and a positive radical value, \(-|c|\) cannot be negative number and must be \(0\).

Plugging in \(c = 0\) into the original equation leaves: 

\(­|a + b - 2x| + \sqrt{(x^2 - a - b)^2} = 0 \)­

As this sum comes to \(0\), both \(|a + b - 2x|\) and \(\sqrt{(x^2 - a - b)^2}\) are equal to \(0\) and therefore equal to one another. This also means that \(|a + b - 2x| = a + b - 2x\) as zero is neither negative nor positive, and that \(\sqrt{(x^2 - a - b)^2} = x^2 - a - b\) as the square and squareroot of zero will always be zero.

Letting them equal one another: 

\(­a + b - 2x =  x^2 - a - b \)­

\(2a + 2b = x^2 + 2x\)

\(2(a+b) = x(x+2)\)

From this one can deduce that \(x = 2\), that \(a = x = 2\) and that \(b = 2\)

If one plugs these values, as well as \(c = 0\), back into \(­|a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} =-|c| \)­ one will get: \(|2+2+0 - 4| + \sqrt{(2^2 - 2 - 2 - 0)^2} = 0\) which holds.

Therefore, \(x + c = 2\)

ANSWER C
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Out of curiosity could it be solved the following way as well?

(i) \(|a + b + c -2x|\) = \(a + b + c - 2x , -a - b - c + 2x\) where we can then check for each case
(ii) \(\sqrt{(x^2 - a - b - c)^2}\) has to be positive so we take \((x^2 - a - b - c)\)
(iii) then combining case 1 of (i) and (ii) we receive: \(a + b + c - 2x + x^2 - a - b - c = -|c|\)
(iv) a, b, c all cancel and now we have \(x^2 - 2x = -|c|\)
(v) move c over, \(x^2 - 2x + |c| = 0\)
(vi) since we know c is positive we use \(a^2 - b^2\): \((x - 1)(x - 1)\)
(vii) which then gives us \(x = 1\) and \(c = 1\)
(viii) \(x + c = 1 + 1 = 2\)
Bunuel
­If \(x\) is a prime number and \(|a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} = - |c|\), what is the value of \(x + c\)?

A. 0
B. 1
C. 2
D. 3
E. Cannot be determined from the given information­

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nickthome
Out of curiosity could it be solved the following way as well?

(i) \(|a + b + c -2x|\) = \(a + b + c - 2x , -a - b - c + 2x\) where we can then check for each case
(ii) \(\sqrt{(x^2 - a - b - c)^2}\) has to be positive so we take \((x^2 - a - b - c)\)
(iii) then combining case 1 of (i) and (ii) we receive: \(a + b + c - 2x + x^2 - a - b - c = -|c|\)
(iv) a, b, c all cancel and now we have \(x^2 - 2x = -|c|\)
(v) move c over, \(x^2 - 2x + |c| = 0\)
(vi) since we know c is positive we use \(a^2 - b^2\): \((x - 1)(x - 1)\)
(vii) which then gives us \(x = 1\) and \(c = 1\)
(viii) \(x + c = 1 + 1 = 2\)


Almost nothing there is correct, unfortunatelly.
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