AccipiterQ wrote:
Bunuel wrote:
If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?
(A) \(\sqrt{2-x}\)
(B) \(2x-6 + \sqrt{2-x}\)
(C) \(\sqrt{2-x} + x-3\)
(D) \(2x-6 + \sqrt{x-2}\)
(E) \(x + \sqrt{x-2}\)
One important note: \(\sqrt{x^2}=|x|\)
\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).
Now, as the expressions under the square roots are more than or equal to zero than \(2-x>0\) --> \(x<2\). Next: as \(x<2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).
\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).
Answer: A.
I don't follow the highlighted part at all. So if you have x-n, and it is supposed to be positive you can just change around the order? Shouldn't the expression just be undefined, since the facts we are given, and the expression we are given can't coexist?
Absolutely not! Ordinarily, if you change the order, it changes everything. But this situation is not ordinary. It is a mod situation i.e. there is modulus around x - n.
Recall how mods are defined:
|x| = x if x is positive
= -x if x is negative
So if you want to get rid of the mod sign, you need to know whether x is positive or negative.
Here we know that x <= 2. For all such values of x, (x-3) is negative.
So |x-3| = -(x - 3) = 3 - x