2000 AIME II Problems/Problem 13
Problem
The equation has exactly two real roots, one of which is , where , and are integers, and are relatively prime, and . Find .
Solution
We may factor the equation as:^{[1]}
Now for real . Thus the real roots must be the roots of the equation . By the quadratic formula the roots of this are:
Thus , and so the final answer is .
^ A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of with half of the polynomial's degree (in this case, divide through by ), and then to use one of the substitutions . In this case, the substitution gives and , which reduces the polynomial to just . Then one can backwards solve for .
Solution 2 (Complex Bash)
It would be really nice if the coefficients were symmetrical. What if we make the substitution, . The the polynomial becomes $\$ (Error compiling LaTeX. ! Missing $ inserted.)-2y^6 - (\frac{i}{\sqrt{10}})y^5 + (\frac{i}{\sqrt{10}})y^3 - (\frac{i}{\sqrt{10}})y - 2\$ It's symmetric! Dividing by and rearranging, we get $\$ (Error compiling LaTeX. ! Missing $ inserted.)-2(y^3 + \frac{1}{y^3}) - (\frac{i}{\sqrt{10}})(y^2 + \frac{1}{y^2}) + (\frac{i}{\sqrt{10}})\$ Now, if we let , we can get the equations $\$ (Error compiling LaTeX. ! Missing $ inserted.)z = y + \frac{1}{y}\$ $z^2 - 2 = y^2 + \frac{1}{y^2}\$ (Error compiling LaTeX. ! Missing $ inserted.)z^3 - 3z = y^3 + \frac{1}{y^3}\$ (These come from squaring and subtracting , then multiplying that result by and subtracting ) $\$ Plugging this into our polynomial, expanding, and rearranging, we get$ (Error compiling LaTeX. ! Missing $ inserted.)\$ $-2z^3 - (\frac{i}{\sqrt{10}})z^2 + 6z + (\frac{3i}{\sqrt{10}})\$ Now, we see that the two$ (Error compiling LaTeX. ! Missing $ inserted.)i0z = \sqrt{3}z = -\sqrt{3}(z - \sqrt{3})(z + \sqrt{3}) = (z^2 - 3)\$ , we know that the other two solutions for z wouldn't result in real solutions for since we have to solve a quadratic with a negative discriminant, then multiply by . We get that . Solving for (using ) we get that , and multiplying this by (because ) we get that for a final answer of
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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