study 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) \(x + \sqrt{x-2}\)
First, \((2 - x)\) has to be \(\ge 0\) for the expression \(\sqrt{2 - x}\) to be defined (GMAT deals only with real numbers). Therefore, \(x \le 2\) .
Even if x <= 2, rt (x-3)^2 is still a valid explanation. Since there is a square below the rt sign the number is obviously positive. How can one conveniently change from (x - 3) to (3 - x)?
\(\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = 3 - x\) (because \(x \le 2\) and thus \(x - 3 \lt 0\) ).
IMO:
\(\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = x - 3\) is valid
Summing up: \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + (x - 3) = (x - 3) + \sqrt{2 - x} + (x - 3) = 2x - 6 + \sqrt{x - 2}\).
M18-16Given: Each expression under the square root is greater than or equal to 0.
Asked: What is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?
\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)
Since all expressions inside square root are non-negative numbers
x^2 - 6x +9 = (x -3)^2 >=0
2- x >=0
x<=2
\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)
\(|x-3| + \sqrt{2 - x} + (x-3)\)
Since x <=2 => x -3 <= -1 <0 => |x-3| = 3-x
\(|x-3| + \sqrt{2 - x} + (x-3)\)
\(3 -x + \sqrt{2 - x} + x-3\)
\(\sqrt{2 - x}\)
IMO A
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Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com