2015 AMC 12B Problems/Problem 12
Contents
Problem
Let , , and be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation ?
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=423
~ pi_is_3.14
Solution 1
The left-hand side of the equation can be factored as , from which it follows that the roots of the equation are , and . The sum of the roots is therefore , and the maximum is achieved by choosing , and . Therefore the answer is
Solution 2
Expand the polynomial. We get
Now, consider a general quadratic equation The two solutions to this are The sum of these roots is
Therefore, reconsidering the polynomial of the problem, the sum of the roots is Now, to maximize this, it is clear that Also, we must have (or vice versa). The reason have to equal these values instead of larger values is because each of is distinct.
Solution 3 (Vieta's Formula)
Expanding gives:
Dividing by the coefficient of the polynomial, 2, and applying Vieta's formulas we know that the sum of the roots of the polynomial is equal to . Obviously, the largest value should occur when b = 9, and due to symmetrically a = 8, and c = 7, or a = 7 and c = 8. Both of these results give the same final answer of (D).
More on Vieta's Formulas: https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas
~PeterDoesPhysics
Solution 3 (Vieta's Formula)
Expanding the formula gives:
Dividing by the coefficient of the polynomial, 2, and applying Vieta's formulas we know that the sum of the roots of the polynomial is equal to . Obviously, the largest value should occur when b = 9, and due to symmetrically a = 8, and c = 7, or a = 7 and c = 8. Both of these results give the same final answer of (D).
More on Vieta's Formulas: https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas
~PeterDoesPhysics
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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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