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Which of the following options always equates to \(\sqrt{9 + x^2 - 6x}\)?
A. x - 3
B. 3 + x
C. |3 - x|
D. |3 + x|
E. 3 - x
A. x - 3
B. 3 + x
C. |3 - x|
D. |3 + x|
E. 3 - x
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19 Feb 2013, 05:09
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Sachin9
Yes, since both |a-b| and |b-a| represent the distance between a and b on the number line.
COMPLETE SOLUTION:
Which of the following options always equates to \(\sqrt{9 + x^2 - 6x}\)?
A. \(x - 3\)
B. \(3 + x\)
C. \(|3 - x|\)
D. \(|3 + x|\)
E. \(3 - x\)
\(\sqrt{9 + x^2 - 6x} = \)
\(=\sqrt{(3 - x)^2} = \)
\(=|3 - x|\)
Note that both \(|x - 3|\) and \(|3 - x|\) represent the distance between \(x\) and 3. Hence, it follows that \(|x - 3| = |3 - x|\).
Answer: C
07 Apr 2015, 10:48
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ankittiss
No, it's totally wrong.
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:
\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;
Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html
All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59
Tips on Numper Properties: number-properties-tips-and-hints-174996.html
All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59
About \(\sqrt{x^2}=|x|\):
Again, the point here is that since square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).
So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?
Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).
So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).
What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).
Hope it helps.
