naeln wrote:
What is \(\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}\) if each term in this expression is correctly defined?
A. \(\sqrt{-x}\)
B. \(1 + \sqrt{x}\)
C. \(1 + \sqrt{-x}\)
D. \(-1 + \sqrt{x}\)
E. \(-1 + \sqrt{-x}\)
Source: OptimusPrep
ALTERNATE METHOD:Hi Readers,
SUBSTITUTION of a correct numbers works very well in such cases but there has to be an observation to be madeSince each term is correctly defined in the expression
\(\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}\)
therefore we must realize that \(\sqrt{-x}\) should be a real number which means that x MUST be a Negative NumberSo let's take X = -4 [I am not taking X = -1 because that may confuse us among options]
Now, \(\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}\) = \(\frac{\sqrt{(-4)^2 -6(-4)+9}}{(-4)-3}+\sqrt{-(-4)}\)
= \(\frac{\sqrt{16 + 24+9}}{-4-3}+\sqrt{(4)}\)
= \(\frac{\sqrt{49}}{-7}+2\)
= \(\frac{7}{-7}+2\)
= \(-1+2\)
= \(1\)
Check Options By Substituting \(X=-4\)
A. \(\sqrt{-x}\) = 2
INCORRECTB. \(1 + \sqrt{x}\) = NOT REAL NUMBER therefore
INCORRECTC. \(1 + \sqrt{-x}\) = \(1+2\) = \(3\)
INCORRECTD. \(-1 + \sqrt{x}\) = NOT REAL NUMBER therefore
INCORRECTE. \(-1 + \sqrt{-x}\) = \(-1 + 2\) = \(+1\)
CORRECTAnswer: Option
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