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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.

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.

For the question to be true. the right side of the equation has to be positive, (hence x has to be smaller than 5) because the left side of the equation is always positive.

s1 tells us that x is a negative number. So, it's sufficient s2 tells us that x is less than 5, so it sufficient.

The question basically wants to know if x<=5 else RHS will be x-5

Statement 1 -x|x|>0 or x|x|<0 (multiply by -1 both sides and reverse the sign) either x<0 or |x|<0 since |x| is always positive or 0 x<0 is true. if x<0 then x<5 hence sufficient

Statement 2 5-x>0 5>x This is what we are looking for hence sufficient

Since L.H.S is a square it will be always positive.and R.H.S 5-x will only be positive when x is less than or equal to 5. so in other words the question can be rephrase as is x<=5

stmt1: -x|x|>0

divide both side by |x| we get -x>0 it means x < 0 sufficient

Let me point out something here: You cannot square both sides to get Is \((\sqrt{(x-5)^2})^2 = (5-x)^2\) ?

People sometimes get confused here. Why can you not square it? It is a question similar to 'Is x = 5?' Can you square both sides here and change the question to 'Is \(x^2 = 25\)?' Please remember, they are not the same. x^2 can be 25 even if x is not 5 ( when x = -5, even then x^2 = 25). Only if it is given to you that x = 5, then you can say that x^2 = 25.

You can rephrase the question in the following manner (and many more ways)

Is \((\sqrt{(x-5)^2}) = (5-x)\) ? Is \(|x-5| = (5-x)\) ? or Is \(|5-x| = (5-x)\)? We know that |x| = x only when x >= 0 So \(|5-x| = (5-x)\) only when 5 - x >= 0 or when x <= 5

Stmnt 1: -x|x| > 0 Since |x| is always positive (or zero), -x must be positive too. So x must be negative. If x < 0, then x is obviously less than 5. Sufficient.

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.

(2) \(5-x>0\) --> \(x<5\). Sufficient.

Answer: D.

Hi Bunuel,

I didn't understand from here : is |x−5|=5−x? --> is x−5≤0? --> is x≤5?.

Modulus always results in an positive value. If x-5>0 then |x−5| should be equal to x-5 However, If x-5<0 then modulus would result the positive value of it i.e. -(x-5)=5-x

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