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

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