Difference between revisions of "2004 AMC 12B Problems/Problem 17"
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<cmath>8(x-x_1)(x-x_2)(x-x_3) = 8x^3 + 4ax^2 + 2bx + a</cmath> | <cmath>8(x-x_1)(x-x_2)(x-x_3) = 8x^3 + 4ax^2 + 2bx + a</cmath> | ||
gives us that <math>a = -8x_1x_2x_3 = -256 \Rightarrow \mathrm{(A)}</math>. | gives us that <math>a = -8x_1x_2x_3 = -256 \Rightarrow \mathrm{(A)}</math>. | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/40mJNmstTEY | ||
== See also == | == See also == | ||
Revision as of 16:25, 3 January 2020
Contents
Problem
For some real numbers 
and 
, the equation
![\[8x^3 + 4ax^2 + 2bx + a = 0\]](http://latex.artofproblemsolving.com/c/7/7/c776fa3cc39e7fb51e32ea87c738b36491d7ed49.png)
has three distinct positive roots. If the sum of the base-
logarithms of the roots is 
, what is the value of ?
Solution
Let the three roots be 
.
![\[\log_2 x_1 + \log_2 x_2 + \log_2 x_3 = \log_2 x_1x_2x_3= 5 \Longrightarrow x_1x_2x_3 = 32\]](http://latex.artofproblemsolving.com/9/e/a/9eaba1d48b6d1b7d491b25a3503629ff84a91bee.png)
By Vieta’s formulas,
![\[8(x-x_1)(x-x_2)(x-x_3) = 8x^3 + 4ax^2 + 2bx + a\]](http://latex.artofproblemsolving.com/0/6/b/06b76979c050b8824bb4a74150704bf5dd0867a8.png)
gives us that 
.
Video Solution
See also
| 2004 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 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. 
