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

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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}=\boxed{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>.
  
 
== See also ==
 
== See also ==

Latest revision as of 11:20, 17 February 2025

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 1

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

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

$x^2=x\sqrt{19}+91$

$x^2-x\sqrt{19}-91 = 0$

$\left.\begin{array}{l}x_1=\frac{\sqrt{19}+\sqrt{383}}{2}\\\\x_2=\frac{\sqrt{19}-\sqrt{383}}{2}\end{array}\right\}A=|x_1|+|x_2|\Rightarrow\sqrt{383}$

$A^2=\boxed{383}$

Solution 2

Let $f(x) = \sqrt{19} + \frac{91}{x}$. Then $x = f(f(f(f(f(x)))))$, from which we realize that $f(x) = x$. This is because if we expand the entire expression, we will get a fraction of the form $\frac{ax + b}{cx + d}$ 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 $f(x)=x$.

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 $A_{}^{2}=\boxed{383}$.

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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