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04 Mar 2020, 02:31
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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
Are You Up For the Challenge: 700 Level Questions
M37-18
A. 1
B. 2
C. 3
D. 4
E. 5
Are You Up For the Challenge: 700 Level Questions
M37-18
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19 Nov 2021, 05:11
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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
04 Mar 2020, 02:54
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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?
\(2=x*(1+\frac{2}{1+\frac{2}{1+...}})\)
--> \(2= x*(1 + x)\)
\(x^{2}+ x-2 =0\)
\((x-1)(x+2)=0\)
--> \(x_1= 1\), \(x_2= -2\)
(x is positive)--> \(x= 1\)
The answer is A.
\(2=x*(1+\frac{2}{1+\frac{2}{1+...}})\)
--> \(2= x*(1 + x)\)
\(x^{2}+ x-2 =0\)
\((x-1)(x+2)=0\)
--> \(x_1= 1\), \(x_2= -2\)
(x is positive)--> \(x= 1\)
The answer is A.
