Difference between revisions of "1951 AHSME Problems/Problem 42"
Yankeesfan (talk | contribs) (Created page with "== Problem == If <math> x =\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}} </math>, then: <math> \textbf{(A)}\ x = 1\qquad\textbf{(B)}\ 0 < x < 1\qquad\textbf{(C)}\ 1 < x < 2\qquad\t...") |
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== Solution == | == Solution == | ||
We note that <math>x^2=1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}=1+x</math>. By the quadratic formula, <math>x=\dfrac{1\pm \sqrt{5}}{2}</math>. Because there are only positive square roots in <math>x</math>, <math>x</math> must be positive, thus, it is <math>\dfrac{1+\sqrt{5}}{2}\approx 1.618</math>. Thus, <math>\boxed{\textbf{(C)}\ 1 < x < 2}</math>. | We note that <math>x^2=1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}=1+x</math>. By the quadratic formula, <math>x=\dfrac{1\pm \sqrt{5}}{2}</math>. Because there are only positive square roots in <math>x</math>, <math>x</math> must be positive, thus, it is <math>\dfrac{1+\sqrt{5}}{2}\approx 1.618</math>. Thus, <math>\boxed{\textbf{(C)}\ 1 < x < 2}</math>. | ||
| − | == See | + | == See Also == |
| − | {{AHSME box|year=1951|num-b=41|num-a=43}} | + | {{AHSME 50p box|year=1951|num-b=41|num-a=43}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 07:59, 29 April 2012
Problem
If 
, then:
Solution
We note that 
. By the quadratic formula, 
. Because there are only positive square roots in 
, must be positive, thus, it is 
. Thus, 
.
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 41 |
Followed by Problem 43 | |
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| All AHSME Problems and Solutions | ||
