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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) \(x + \sqrt{x-2}\)
M18-16
(A) \(\sqrt{2-x}\)
(B) \(2x-6 + \sqrt{2-x}\)
(C) \(\sqrt{2-x} + x-3\)
(D) \(2x-6 + \sqrt{x-2}\)
(E) \(x + \sqrt{x-2}\)
M18-16
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09 Oct 2013, 02:22
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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. \(x + \sqrt{x - 2}\)
Absolute Value Properties:
When \(x \le 0\), then \(|x|=-x\), or more generally, when \(\text{some expression} \le 0\), then \(|\text{some expression}| = -(\text{some expression})\).
When \(x \ge 0\), then \(|x|=x\), or more generally, when \(\text{some expression} \ge 0\), then \(|\text{some expression}| = \text{some expression}\).
Another important property to remember is \(\sqrt{x^2}=|x|\).
Back to the Original Question:
Now, since the expressions under the square roots are greater than or equal to zero, then \(2-x \ge 0\), which implies \(x \le 2\). Next, since \(x \le 2\), the expression \(x-3\) inside the absolute value is negative. Therefore, \(|x-3|=-(x-3)=-x+3\). This allows us to rewrite the above expression as follows:
Answer: A
A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)
Absolute Value Properties:
When \(x \le 0\), then \(|x|=-x\), or more generally, when \(\text{some expression} \le 0\), then \(|\text{some expression}| = -(\text{some expression})\).
When \(x \ge 0\), then \(|x|=x\), or more generally, when \(\text{some expression} \ge 0\), then \(|\text{some expression}| = \text{some expression}\).
Another important property to remember is \(\sqrt{x^2}=|x|\).
Back to the Original Question:
- \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\)
\(=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=\)
\(=|x-3|+\sqrt{2-x}+x-3\).
Now, since the expressions under the square roots are greater than or equal to zero, then \(2-x \ge 0\), which implies \(x \le 2\). Next, since \(x \le 2\), the expression \(x-3\) inside the absolute value is negative. Therefore, \(|x-3|=-(x-3)=-x+3\). This allows us to rewrite the above expression as follows:
- \(|x-3|+\sqrt{2-x}+x-3=\)
\(=-x+3+\sqrt{2-x}+x-3=\)
\(=\sqrt{2-x}\).
Answer: A
08 Oct 2013, 21:48
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jlgdr
The correct answer will be (A)
The expression becomes
\(|x-3| + \sqrt{2 - x} + (x - 3)\)
Note that \(\sqrt{x^2} = |x|\) (and not x)
Since we know that x < 2, (x - 3) must be negative. Hence \(|x-3| = -(x - 3)\)
Expression becomes: \(-(x - 3) + \sqrt{2 - x} + (x - 3) = \sqrt{2 - x}\)
