1991 AIME Problems/Problem 7

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

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
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1991 AIME Problems/Problem 7

Contents

Problem

Solution 1

Solution 2

See also