Difference between revisions of "1991 AIME Problems/Problem 7"
m (→Solution) |
Hashtagmath (talk | contribs) m (→Solution) |
||
Line 7: | Line 7: | ||
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>. | 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>. | ||
− | 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>. | + | 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>. |
== Solution 2 == | == Solution 2 == |
Revision as of 00:03, 8 January 2019
Contents
Problem
Find , where
is the sum of the absolute values of all roots of the following equation:
![$x = \sqrt{19} + \frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{x}}}}}}}}}$](http://latex.artofproblemsolving.com/b/0/2/b02f23aa1a41aa56a1e00e986776128fffe43321.png)
Solution
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
.
Solution 2
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.