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If \(x\) is a positive number and \(\frac{2}{1+\frac{2}{1+\frac{2}{1+...}}}=x\), where the given expression extends to an infinite number of fractions, then what is the value of \(x\)?
A. \(1\)
B. \(2\)
C. \(3\)
D. \(4\)
E. \(5\)
A. \(1\)
B. \(2\)
C. \(3\)
D. \(4\)
E. \(5\)
19 Nov 2021, 05:10
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Official Solution:
If \(x\) is a positive number and \(\frac{2}{1+\frac{2}{1+\frac{2}{1+...}}}=x\), where the given expression extends to an infinite number of fractions, then what is the value of \(x\)?
A. \(1\)
B. \(2\)
C. \(3\)
D. \(4\)
E. \(5\)
Given: \(\frac{2}{1+\frac{2}{1+\frac{2}{1+...}}}=x\);
\(\frac{2}{1+(\frac{2}{1+\frac{2}{1+...}})}=x\).
Since the expression in the denominator extends infinitely then expression in brackets would equal to \(x\) itself and we can safely replace it with \(x\) and rewrite the given expression as:
\(\frac{2}{1+x}=x\);
\(2=x+x^2;\)
\((x - 1)(x + 2) = 0\)
\(x=1\) or \(x=-2\). Since given that \(x\) is a positive number, then \(x=1\) only.
Answer: A
If \(x\) is a positive number and \(\frac{2}{1+\frac{2}{1+\frac{2}{1+...}}}=x\), where the given expression extends to an infinite number of fractions, then what is the value of \(x\)?
A. \(1\)
B. \(2\)
C. \(3\)
D. \(4\)
E. \(5\)
Given: \(\frac{2}{1+\frac{2}{1+\frac{2}{1+...}}}=x\);
\(\frac{2}{1+(\frac{2}{1+\frac{2}{1+...}})}=x\).
Since the expression in the denominator extends infinitely then expression in brackets would equal to \(x\) itself and we can safely replace it with \(x\) and rewrite the given expression as:
\(\frac{2}{1+x}=x\);
\(2=x+x^2;\)
\((x - 1)(x + 2) = 0\)
\(x=1\) or \(x=-2\). Since given that \(x\) is a positive number, then \(x=1\) only.
Answer: A
17 Oct 2024, 07:27
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Bunuel
How can we safely assume that the expression can be equal to x?
What is this concept? Can you please help? Bunuel
