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.
P.S. Explanation of: \(\sqrt{x^2}=|x|\).
The point here is that as
square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\).
So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?
Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).
So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).
What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).
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