Difference between revisions of "1996 AIME Problems/Problem 5"
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== Problem == | == Problem == | ||
| − | Suppose that the | + | Suppose that the [[root]]s of <math>x^3+3x^2+4x-11=0</math> are <math>a</math>, <math>b</math>, and <math>c</math>, and that the roots of <math>x^3+rx^2+sx+t=0</math> are <math>a+b</math>, <math>b+c</math>, and <math>c+a</math>. Find <math>t</math>. |
== Solution == | == Solution == | ||
| − | { | + | Use [[Vieta's formulas]]. These tell us that <math>a + b + c = s = -3</math>, <math>ab + bc + ca = 4</math>, and <math>abc = 11</math> for the first polynomial. Then: |
| + | <div style="text-align:center;"> | ||
| + | <math>\begin{eqnarray*} | ||
| + | t &=& -(a+b)(b+c)(c+a)\\ | ||
| + | &=& -[(s-a)(s-b)(s-c)]\\ | ||
| + | &=& -[(-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</math></div> | ||
| + | |||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1996|num-b=4|num-a=6}} | |
| − | + | [[Category:Intermediate Algebra Problems]] | |
Revision as of 19:07, 24 September 2007
Problem
Suppose that the roots of 
are 
, 
, and 
, and that the roots of 
are 
, 
, and 
. Find 
.
Solution
Use Vieta's formulas. These tell us that 
, 
, and 
for the first polynomial. Then:
$\begin{eqnarray*} t &=& -(a+b)(b+c)(c+a)\\
&=& -[(s-a)(s-b)(s-c)]\\ &=& -[(-3-a)(-3-b)(-3-c)]\\ &=& [(a+3)(b+3)(c+3)]\\ &=& abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\t &=& 11 + 3(4) + 9(-3) + 27 = 23$ (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 | ||
