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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.

Is root(x-5)^2 = 5 - x ? [#permalink]
09 Sep 2009, 14:41

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dhushan wrote:

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.

Is \(\sqrt{(x-5)^2}=5-x\)?

First of all, recall that \(\sqrt{x^2}=|x|\).

Is \(\sqrt{(x-5)^2}=5-x\)? --> is \(|x-5|=5-x\)? --> is \(x-5\leq{0}\)? --> is \(x\leq{5}\)?

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

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