Carcass 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) \(x + \sqrt{x-2}\)
The goal here is to determine which expression is equivalent to \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\).
STRATEGY: Upon reading any GRE Multiple Choice question, we should always ask, Can I use the answer choices to my advantage?
In this case, we can easily plug in values of x to test for equivalency.
Now let's give ourselves up to 20 seconds to identify a faster approach.
In this case, we can also apply some algebraic factoring techniques to simplify the given expression.
Since the algebraic approach seems both tricky and time consuming, I'm going to test for equivalency.Key concept: If two expressions are equivalent, they must evaluate to the same value for every possible value of x.
For example, since the expression 2x + 3x is equivalent to the expression 5x, the two expressions will evaluate to the same number for every value of x.
So, if x = 7, the expression 2x + 3x = 2(7) + 3(7) = 14 + 21 = 35, and the expression 5x = 5(7) = 35Let's evaluate the given expression for \(x = 0\).
We get: \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3 = \sqrt{0^2 - 6(0) + 9} + \sqrt{2 - 0} + 0 - 3\)
\(= \sqrt{9} + \sqrt{2} - 3\)
\(= 3 + \sqrt{2} - 3\)
\(= \sqrt{2}\)
So, when \(x = 0\), the value of the given expression is \(\sqrt{2}\)
We'll now evaluate each answer choice for \(x = 0\) and eliminate those that don't evaluate to \(\sqrt{2}\)
(A) \(\sqrt{2-x}=\sqrt{2-0}=\sqrt{2}\)
KEEP(B) \(2x-6 + \sqrt{2-x} = 2(0) -6 + \sqrt{2-0} = -6 + \sqrt{2}\). ELIMINATE
(C) \(\sqrt{2-x} + x-3=\sqrt{2-0} + 0-3=\sqrt{2} -3\). ELIMINATE
(D) \(2x-6 + \sqrt{x-2} = 2(0) -6 + \sqrt{0-2} = -6 + \sqrt{-2} =\) no defined value. ELIMINATE
(E) \(x + \sqrt{x-2}= 0 + \sqrt{0-2} = \sqrt{-2} =\) no defined value. ELIMINATE
The process of elimination, the correct answer is A.