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Difference between revisions of "1991 AIME Problems/Problem 7"

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== Problem ==
 
== Problem ==
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Find <math>A^2_{}</math>, where <math>A^{}_{}</math> is the sum of the absolute values of all roots of the following equation:
+
Find <math>A^2_{}</math>, where <math>A^{}_{}</math> is the sum of the [[absolute value]]s of all roots of the following equation:
−
<div style="text-align:center"><math>x = \sqrt{19} + \frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{x}}}}}}}}}
+
<div style="text-align:center"><math>x = \sqrt{19} + \frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{x}}}}}}}}}
 
</math></div>
 
</math></div>
  
 
== Solution ==
 
== Solution ==
−
The given finite expansion can be easily seen that reduces to solve the quadratic equation <math>x_{}^{2}-\sqrt{19}x-91=0</math>. The solutions are <math>x_{\pm}^{}=</math><math>\frac{\sqrt{19}\pm\sqrt{383}}{2}</math>. Therefore, <math>A_{}^{}=\vert x_{+}\vert+\vert x_{-}\vert=\sqrt{383}</math>. We conclude that <math>A_{}^{2}=383</math>.
+
Let <math>f(x) = \sqrt{19} + \frac{91}{x}</math>. Then <math>x = f(f(f(f(f(x)))))</math>, from which we hypothesize that <math>f(x) = x</math>. The given finite expansion can then be easily seen to reduce to the [[quadratic equation]] <math>x_{}^{2}-\sqrt{19}x-91=0</math>. The solutions are <math>x_{\pm}^{}=</math><math>\frac{\sqrt{19}\pm\sqrt{383}}{2}</math>. Therefore, <math>A_{}^{}=\vert x_{+}\vert+\vert x_{-}\vert=\sqrt{383}</math>. We conclude that <math>A_{}^{2}=383</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=6|num-a=8}}
 
{{AIME box|year=1991|num-b=6|num-a=8}}
 +
 +
[[Category:Intermediate Algebra Problems]]

Revision as of 18:16, 22 October 2007

Problem

Find $A^2_{}$, where $A^{}_{}$ is the sum of the absolute values of all roots of the following equation:

$x = \sqrt{19} + \frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{x}}}}}}}}}$

Solution

Let $f(x) = \sqrt{19} + \frac{91}{x}$. Then $x = f(f(f(f(f(x)))))$, from which we hypothesize that $f(x) = x$. The given finite expansion can then be easily seen to reduce to the quadratic equation $x_{}^{2}-\sqrt{19}x-91=0$. The solutions are $x_{\pm}^{}=$$\frac{\sqrt{19}\pm\sqrt{383}}{2}$. Therefore, $A_{}^{}=\vert x_{+}\vert+\vert x_{-}\vert=\sqrt{383}$. We conclude that $A_{}^{2}=383$.

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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