Difference between revisions of "2015 AMC 12B Problems/Problem 12"
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<math>\textbf{(A)}\; 15 \qquad\textbf{(B)}\; 15.5 \qquad\textbf{(C)}\; 16 \qquad\textbf{(D)} 16.5 \qquad\textbf{(E)}\; 17</math> | <math>\textbf{(A)}\; 15 \qquad\textbf{(B)}\; 15.5 \qquad\textbf{(C)}\; 16 \qquad\textbf{(D)} 16.5 \qquad\textbf{(E)}\; 17</math> | ||
| − | ==Solution== | + | == 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 <math>(x-b)(x-a+x-c) = (x-b)(2x-(a+c))</math>, from which it follows that the roots of the equation are <math>x=b</math>, and <math>x=\tfrac{a+c}{2}</math>. The sum of the roots is therefore <math>b + \tfrac{a+c}{2}</math>, and the maximum is achieved by choosing <math>b=9</math>, and <math>\{a,c\}=\{7,8\}</math>. Therefore the answer is <math>9 + \tfrac{7+8}{2} = 9 + 7.5 = \boxed{\textbf{(D)}\; 16.5}.</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | Expand the polynomial. We get <math>(x-a)(x-b)+(x-b)(x-c)=x^2-(a+b)x+ab+x^2-(b+c)x+bc=2x^2-(a+2b+c)x+(ab+bc).</math> | ||
| + | |||
| + | Now, consider a general quadratic equation <math>ax^2+bx+c=0.</math> The two solutions to this are <cmath>\dfrac{-b}{2a}+\dfrac{\sqrt{b^2-4ac}}{2a}, \dfrac{-b}{2a}-\dfrac{\sqrt{b^2-4ac}}{2a}.</cmath> The sum of these roots is <cmath>\dfrac{-b}{a}.</cmath> | ||
| + | |||
| + | Therefore, reconsidering the polynomial of the problem, the sum of the roots is <cmath>\dfrac{a+2b+c}{2}.</cmath> Now, to maximize this, it is clear that <math>b=9.</math> Also, we must have <math>a=8, c=7</math> (or vice versa). The reason <math>a,c</math> have to equal these values instead of larger values is because each of <math>a,b,c</math> is distinct. | ||
| + | |||
| + | ==Solution 3 (Vieta's Formula)== | ||
| + | Expanding gives: <math>2x^2-(a+2b+c)x+(ab+bc).</math> | ||
| + | |||
| + | 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 <math>\frac{a + 2b + c}{2}</math>. 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: <math>2x^2-(a+2b+c)x+(ab+bc).</math> | ||
| + | |||
| + | 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 <math>\frac{a + 2b + c}{2}</math>. 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== | ==See Also== | ||
{{AMC12 box|year=2015|ab=B|num-a=13|num-b=11}} | {{AMC12 box|year=2015|ab=B|num-a=13|num-b=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 23:02, 24 August 2024
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 ![\[\dfrac{-b}{2a}+\dfrac{\sqrt{b^2-4ac}}{2a}, \dfrac{-b}{2a}-\dfrac{\sqrt{b^2-4ac}}{2a}.\]](http://latex.artofproblemsolving.com/c/2/4/c24d41bb4f10c62d76b27486cf8aa36a47dec169.png)
The sum of these roots is ![\[\dfrac{-b}{a}.\]](http://latex.artofproblemsolving.com/1/0/e/10ec902f7062667b933d2f9894b5c2bfd9eea4b1.png)
Therefore, reconsidering the polynomial of the problem, the sum of the roots is ![\[\dfrac{a+2b+c}{2}.\]](http://latex.artofproblemsolving.com/3/9/8/39806a75adbb86c84bd995a73020aee4c5f15b33.png)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
