siddharthsinha123 wrote:
If x<0, what is the value of \(\sqrt[4]{(x-3)^4} + \sqrt{-x|x|}\)
A. -3
B. x+3
C. 3-2x
D. 2x-3
E. 2
Two things to remember: (1) \(\sqrt{x^2}=|x|\)
(2) When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);
When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\).
Back to the question:According to (1) \(\sqrt[4]{(x-3)^4}=|x-3|\). Now, since it's given that x is negative, then x - 3 will also be negative, thus according to (2) \(|x-3| = -(x-3)=3-x\).
Next, again since x is negative, then \(|x|=-x\). Thus, \(\sqrt{-x|x|}=\sqrt{(-x)(-x)}=\sqrt{x^2}=|x|=-x\).
So, finally we have \(\sqrt[4]{(x-3)^4} + \sqrt{-x|x|}=3-x-x=3-2x\).
Answer: C.
When it's written |x|, don't we always take the positive value no matter what is inside. Why are we even considering the negative value of x when absolute value of a number is positive. Please explain.