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Gemelo90
I was testing different numbers and came to the conclusion that B always has to be true.
a-b is always greater than 0. Since it involves a square root of a square which is always a positive or a zero.
So a-b > 0
a>b.
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If \(a-b > \sqrt{(b+c-a)^2}\) which of the following must be true?


A. a>0

B. a<b

C. b=0

D. c>0

E. c<0

Given : \(a-b > \sqrt{(b+c-a)^2}\)

We know that \(\sqrt{(x)^2}\) = \(|x|\)

So, \(a - b > |b+c-a|\)

Therefore we have 2 cases :

Case 1 :

\(a - b > b + c - a\)

\(2a - 2b > c\)

\(a - b > \frac{c}{2}\) (No match)

Case 2:

\(a - b > -(b + c - a)\)

\(a - b > - b - c + a\)

\(a - a - b + b > - c\)

\(0 > - c\)

\(c > 0\)

(D)
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Another Approach:
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Another approach:
(a-b) > 0 because square root is always non negative.
therefore, (b-a) < 0
hence c must be >0, so that (b+c-a) is non negative.
(If (b+c-a) is negative, square root is not defined)

Hence, c > 0
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