Difference between revisions of "2014 AMC 12B Problems/Problem 11"
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The line through <math>Q</math> only meets the parabola if the system of equations <math>y = x^2</math> and <math>y = m(x - 20) + 14</math> has one or more solution(s) (the second equation comes from point-slope form). We can equate <math>y</math> and get the equation <math>x^2 = mx - 20m + 14</math>. Now if we move all the terms to the left, we get <math>x^2 - mx - (20m - 14) = 0</math>. We want this equation to have no solutions, and are trying to find the bounds of <math>r</math> and <math>s</math> for <math>m</math> such that this is unsolvable. A quadratic is unsolvable across the reals if its discriminant is less than 0. So, we set the discriminant to be less than 0: | The line through <math>Q</math> only meets the parabola if the system of equations <math>y = x^2</math> and <math>y = m(x - 20) + 14</math> has one or more solution(s) (the second equation comes from point-slope form). We can equate <math>y</math> and get the equation <math>x^2 = mx - 20m + 14</math>. Now if we move all the terms to the left, we get <math>x^2 - mx - (20m - 14) = 0</math>. We want this equation to have no solutions, and are trying to find the bounds of <math>r</math> and <math>s</math> for <math>m</math> such that this is unsolvable. A quadratic is unsolvable across the reals if its discriminant is less than 0. So, we set the discriminant to be less than 0: | ||
<cmath>\sqrt{m^2 - 80m - 56} \le 0</cmath> | <cmath>\sqrt{m^2 - 80m - 56} \le 0</cmath> | ||
− | We can square each side, keeping track of extraneous solutions: the resulting equation is not just <math>m^2 - 80m - 14 = 0</math>, but <math>\pm(m^2 - 80m - 66) = 0</math>. The solutions to this equation are <math>r</math> and <math>s</math>: to get their sum we use Vieta's Formulas and get <math>r + s = \pm80</math>. Because <math>-80</math> is not a valid answer choice, we can know for certain that the correct answer is <math>\boxed{(\ | + | We can square each side, keeping track of extraneous solutions: the resulting equation is not just <math>m^2 - 80m - 14 = 0</math>, but <math>\pm(m^2 - 80m - 66) = 0</math>. The solutions to this equation are <math>r</math> and <math>s</math>: to get their sum we use Vieta's Formulas and get <math>r + s = \pm80</math>. Because <math>-80</math> is not a valid answer choice, we can know for certain that the correct answer is <math>\boxed{(\textbf{E})\:80}</math>. |
Revision as of 19:20, 20 February 2014
Problem
Let be the parabola with equation and let . There are real numbers and such that the line through with slope does not intersect if and only if . What is ?
\[\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}}\ 52\qquad\textbf{(E)}\ 80\] (Error compiling LaTeX. Unknown error_msg)
Solution
The line through only meets the parabola if the system of equations and has one or more solution(s) (the second equation comes from point-slope form). We can equate and get the equation . Now if we move all the terms to the left, we get . We want this equation to have no solutions, and are trying to find the bounds of and for such that this is unsolvable. A quadratic is unsolvable across the reals if its discriminant is less than 0. So, we set the discriminant to be less than 0: We can square each side, keeping track of extraneous solutions: the resulting equation is not just , but . The solutions to this equation are and : to get their sum we use Vieta's Formulas and get . Because is not a valid answer choice, we can know for certain that the correct answer is .