Difference between revisions of "1996 AIME Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | + | By [[Vieta's formulas]] on the polynomial <math>P(x) = x^3+3x^2+4x-11 = (x-a)(x-b)(x-c) = 0</math>, we have <math>a + b + c = s = -3</math>, <math>ab + bc + ca = 4</math>, and <math>abc = 11</math>. Then | |
| − | < | + | <center><math>t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)</math></center> |
| − | <math> | + | This is just the definition for <math>-P(-3) = \boxed{023}</math>. |
| − | t | + | |
| − | + | Alternatively, we can expand the expression to get | |
| − | + | <center><math>\begin{align*} | |
| − | &= | + | t &= -(-3-a)(-3-b)(-3-c)\\ |
| − | &= | + | &= (a+3)(b+3)(c+3)\\ |
| − | t &= | + | &= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\ |
| + | t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}</math></center> | ||
== See also == | == See also == | ||
Revision as of 16:18, 21 April 2008
Problem
Suppose that the roots of 
are 
, 
, and 
, and that the roots of 
are 
, 
, and 
. Find 
.
Solution
By Vieta's formulas on the polynomial 
, we have 
, 
, and 
. Then
This is just the definition for 
.
Alternatively, we can expand the expression to get
t &= -(-3-a)(-3-b)(-3-c)\\
&= (a+3)(b+3)(c+3)\\ &= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
See also
| 1996 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
