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| − | ==Problem 4 ==
| + | #redirect [[2022 AMC 10B Problems/Problem 7]] |
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| − | For how many values of the constant <math>k</math> will the polynomial <math>x^{2}+kx+36</math> have two distinct integer roots?
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| − | <math>\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }14 \qquad \textbf{(E) }16</math>
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| − | == Solution ==
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| − | Using Vieta's, denote the two distinct integer roots as <math>r_1,r_2</math>. Then
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| − | <cmath>r_1 \cdot r_2 = 36</cmath>
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| − | We list out the factors of <math>36</math> as follows: <cmath>\{1,2,3,4,6,9,12,18,36\}</cmath>
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| − | Then, we notice that there are exactly 4 pairings of distinct integers that multiply to <math>36</math> (everything excluding <math>6</math>).
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| − | Their corresponding <math>k = -(r_1 + r_2)</math> values are also distinct, which ensures we aren't double counting any pairs.
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| − | However, we aren't done yet! <math>r_1</math> and <math>r_2</math> can be negative. Since the signs of <math>r_1</math> and <math>r_2</math> must match and being negative doesn't change the number of options, we simply double the number of positive pairings to arrive at <math>\boxed{\textbf{(B)} \ 8}</math>
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| − | ~ <math>\color{magenta} zoomanTV</math>
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| − | ==See Also==
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| − | {{AMC12 box|year=2022|ab=B|num-b=3|num-a=5}}
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| − | {{MAA Notice}}
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