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| − | == Problem ==
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| − | Let <math>\mathcal P</math> be the parabola with equation <math>y = x^2</math> and let <math>Q = (20, 14)</math>. There are real numbers <math>r</math> and <math>s</math> such that the line through <math>Q</math> with slope <math>m</math> does not intersect <math>\mathcal P</math> if and only if <math>r < m < s</math>. What is <math>r + s</math>?
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| − | <cmath> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}}\ 52\qquad\textbf{(E)}\ 80 </cmath>
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| − | == Solution ==
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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:
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| − | <cmath>\sqrt{m^2 - 80m - 56} \le 0</cmath>
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| − | We can square each side to get <math>m^2 - 80m - 56 = 0</math>. Because <math>r</math> and <math>s</math> are the solutions to this equation, we can use Vieta's Formulas to find the sum: <math>r + s = -\frac{-80}{1} = \boxed{80}</math>, making the correct answer <math>\textbf{(E)}\:80</math>.
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