Difference between revisions of "2017 AMC 12B Problems/Problem 23"
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<cmath> \frac{9a-1}{a} = 2 + 3 + x_1= 3 + 4 + x_2 = 2 + 4 + x_3</cmath> | <cmath> \frac{9a-1}{a} = 2 + 3 + x_1= 3 + 4 + x_2 = 2 + 4 + x_3</cmath> | ||
Adding all three equations up, we get | Adding all three equations up, we get | ||
− | <cmath> 3(\frac{9a-1}{a}) = 18 + x_1 + x_2 + x_3 = 18 + 24</cmath> | + | <cmath> 3\left(\frac{9a-1}{a}\right) = 18 + x_1 + x_2 + x_3 = 18 + 24</cmath> |
Solving this equation, we get <math>a = -\frac{1}{5}</math>. We finish as Solution 1 does. | Solving this equation, we get <math>a = -\frac{1}{5}</math>. We finish as Solution 1 does. | ||
<math>\boxed{\textbf{(D)}\frac{24}{5}}</math>. | <math>\boxed{\textbf{(D)}\frac{24}{5}}</math>. |
Revision as of 22:25, 4 November 2021
Contents
[hide]Problem
The graph of , where is a polynomial of degree , contains points , , and . Lines , , and intersect the graph again at points , , and , respectively, and the sum of the -coordinates of , , and is 24. What is ?
Solution
First, we can define , which contains points , , and . Now we find that lines , , and are defined by the equations , , and respectively. Since we want to find the -coordinates of the intersections of these lines and , we set each of them to , and synthetically divide by the solutions we already know exist (eg. if we were looking at line , we would synthetically divide by the solutions and , because we already know intersects the graph at and , which have -coordinates of and ). After completing this process on all three lines, we get that the -coordinates of , , and are , , and respectively. Adding these together, we get which gives us . Substituting this back into the original equation, we get , and
Solution by vedadehhc
Solution 2
First of all, . Let's say the line is , and is the coordinate of the third intersection, then , , and are the three roots of . The values of and have no effect on the sum of the 3 roots, because the coefficient of the term is always . So we have Adding all three equations up, we get Solving this equation, we get . We finish as Solution 1 does. .
- Mathdummy
Cleaned up by SSding
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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