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16 Sep 2014, 01:24
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B
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Difficulty:
35%
(medium)
Question Stats:
61% (00:53) correct
39%
(01:32)
wrong
based on 139
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If \(x\) is a positive number and is equal to \(\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...} } } } }\), where the given expression extends to an infinite number of roots, then what is the value of \(x\)?
A. \(\sqrt{6}\)
B. 3
C. \(1+\sqrt{6}\)
D. \(2\sqrt{3}\)
E. 6
A. \(\sqrt{6}\)
B. 3
C. \(1+\sqrt{6}\)
D. \(2\sqrt{3}\)
E. 6
16 Sep 2014, 01:24
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Official Solution:
If \(x\) is a positive number and is equal to \(\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...} } } } }\), where the given expression extends to an infinite number of roots, then what is the value of \(x\)?
A. \(\sqrt{6}\)
B. 3
C. \(1+\sqrt{6}\)
D. \(2\sqrt{3}\)
E. 6
Given: \(x \gt 0\) and \(x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...} } } } }\).
Re-write it as: \(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...} ) } } } }\). As the expression under the square root extends infinitely, the expression in brackets would equal to \(x\) itself. Therefore, we can safely replace it with \(x\) and rewrite the given expression as \(x=\sqrt{6+x}\). Square both sides to get: \(x^2=6+x\). Now, re-arrange and factorize to obtain \((x+2)(x-3)=0\). This gives us \(x=-2\) or \(x=3\). But since \(x \gt 0\), the valid solution is: \(x=3\).
Answer: B
If \(x\) is a positive number and is equal to \(\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...} } } } }\), where the given expression extends to an infinite number of roots, then what is the value of \(x\)?
A. \(\sqrt{6}\)
B. 3
C. \(1+\sqrt{6}\)
D. \(2\sqrt{3}\)
E. 6
Given: \(x \gt 0\) and \(x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...} } } } }\).
Re-write it as: \(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...} ) } } } }\). As the expression under the square root extends infinitely, the expression in brackets would equal to \(x\) itself. Therefore, we can safely replace it with \(x\) and rewrite the given expression as \(x=\sqrt{6+x}\). Square both sides to get: \(x^2=6+x\). Now, re-arrange and factorize to obtain \((x+2)(x-3)=0\). This gives us \(x=-2\) or \(x=3\). But since \(x \gt 0\), the valid solution is: \(x=3\).
Answer: B
General Discussion
31 Aug 2016, 04:41
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I think this is a high-quality question and I agree with explanation.
