Bunuel wrote:

metallicafan wrote:

Is sqrt(x-5)^2 = 5-x?

(1) -xlxl > 0

(2) 5-x > 0

PS. Is always this true?: sqrt{x^2} = lxl ?

Is

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

Remember:

\sqrt{x^2}=|x|.

So "is

\sqrt{(x-5)^2}=5-x?" becomes: is

|x-5|=5-x?

|x-5|=5-x is true only for

x\leq{5}, as in this case

\{LHS=|x-5|=5-x\}=\{RHS=5-x\}. So we have that if

x\leq{5}, then

|x-5|=5-x is true.

Basically question asks is

x\leq{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, so

x is less than 5 too. Sufficient.

(2)

5-x>0 -->

x<5. Sufficient.

Answer: D.

Hope it helps.

Hi Bunuel,

I get confused in the following concept. Can you please help me:

-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, so

x is less than 5 too

Can't we solve it as follows:

As

|x| is never negative ->

-x|x| > 0= -x*x = -x^2 >0 = x^2<0 (multiplying by -ve sign and flipping sign)

x^2<0 =>

sqrt{x^2} <0=> lxl <0 (as

sqrt{x^2} =lxl )

Since lxl cannot be negative and lxl <0 that implies X<0

I have reached to same conclusion as yours but wanted to confirm if my approach is right. Please explain.Thanks

This approach is not right.

The red parts are not correct.

.