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Just want to make sure that my line of thinking is right. I got D as well.

sqrt(x-5)^2 = 5-x ==> |x-5| = 5-x

which means that if

x-5<0, -(x-5) = 5-x = no solution, therefore x<5

x-5>0, x-5 = 5-x, which means that x>5 and x = 5

from (1) -x|x| > 0, we know that x has to be negative -(-x)|-x| is the only way to get a number greater than 0. Therefore, this means that of the three possible solutions for x, only this x<5 hold true.

from (2) 5-x>0, therefore x<5.

Can someone please point out if there is something wrong with my reasoning.

Just want to make sure that my line of thinking is right. I got D as well.

sqrt(x-5)^2 = 5-x ==> |x-5| = 5-x

which means that if

x-5<0, -(x-5) = 5-x = no solution, therefore x<5

x-5>0, x-5 = 5-x, which means that x>5 and x = 5

from (1) -x|x| > 0, we know that x has to be negative -(-x)|-x| is the only way to get a number greater than 0. Therefore, this means that of the three possible solutions for x, only this x<5 hold true.

from (2) 5-x>0, therefore x<5.

Can someone please point out if there is something wrong with my reasoning.

\sqrt{(x-5)^2}=5-x (|x-5| = 5-x) have solution only when 5-x>=0 --> x<=5, because \sqrt{(x-5)^2} (or as you wrote |x-5|) is never negative, therefore for \sqrt{(x-5)^2}=5-x to be true 5-x must be >=0, so here x<=5.

(1) -x|x| > 0 --> |x| is never negative (positive or zero), so in order to have -x|x| > 0, -x must be positive -x>0 --> x<0. Sufficient.