Difference between revisions of "1991 AIME Problems/Problem 7"
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== Problem == | == Problem == | ||
| − | Find <math>A^2_{}</math>, where <math>A^{}_{}</math> is the sum of the [[absolute value]]s of all roots of the following equation: | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>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}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\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><!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
| − | == Solution == | + | == Solution 1 == |
| − | + | <math>x=\sqrt{19}+\underbrace{\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{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>. | + | <math>x=\sqrt{19}+\frac{91}{x}</math> |
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| + | <math>x^2=x\sqrt{19}+91</math> | ||
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| + | <math>x^2-x\sqrt{19}-91 = 0</math> | ||
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| + | <math>\left.\begin{array}{l}x_1=\frac{\sqrt{19}+\sqrt{383}}{2}\\\\x_2=\frac{\sqrt{19}-\sqrt{383}}{2}\end{array}\right\}A=|x_1|+|x_2|\Rightarrow\sqrt{383}</math> | ||
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| + | <math>A^2=\boxed{383}</math> | ||
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| + | == Solution 2 == | ||
| + | Let <math>f(x) = \sqrt{19} + \frac{91}{x}</math>. Then <math>x = f(f(f(f(f(x)))))</math>, from which we realize that <math>f(x) = x</math>. This is because if we expand the entire expression, we will get a fraction of the form <math>\frac{ax + b}{cx + d}</math> on the right hand side, which makes the equation simplify to a quadratic. As this quadratic will have two roots, they must be the same roots as the quadratic <math>f(x)=x</math>. | ||
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| + | 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}=\boxed{383}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 04:15, 13 June 2022
Contents
Problem
Find 
, where 
is the sum of the absolute values of all roots of the following equation:
Solution 1
Solution 2
Let 
. Then 
, from which we realize that 
. This is because if we expand the entire expression, we will get a fraction of the form 
on the right hand side, which makes the equation simplify to a quadratic. As this quadratic will have two roots, they must be the same roots as the quadratic 
.
The given finite expansion can then be easily seen to reduce to the quadratic equation 
. The solutions are 
. Therefore, 
. We conclude that 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
