Difference between revisions of "1991 AIME Problems/Problem 7"
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<math>x^2=x\sqrt{19}+91</math> | <math>x^2=x\sqrt{19}+91</math> | ||
− | <math>x^2-x\sqrt{19}-91</math> | + | <math>x^2-x\sqrt{19}-91 = 0</math> |
<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> | <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> |
Revision as of 15:05, 20 February 2019
Contents
Problem
Find , where is the sum of the absolute values of all roots of the following equation:
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.