# 2004 AMC 12B Problems/Problem 17

## Problem

For some real numbers $a$ and $b$, the equation $$8x^3 + 4ax^2 + 2bx + a = 0$$ has three distinct positive roots. If the sum of the base- $2$ logarithms of the roots is $5$, what is the value of $a$? $\mathrm{(A)}\ -256 \qquad\mathrm{(B)}\ -64 \qquad\mathrm{(C)}\ -8 \qquad\mathrm{(D)}\ 64 \qquad\mathrm{(E)}\ 256$

## Solution

Let the three roots be $x_1,x_2,x_3$. $$\log_2 x_1 + \log_2 x_2 + \log_2 x_3 = \log_2 x_1x_2x_3= 5 \Longrightarrow x_1x_2x_3 = 32$$ By Vieta’s formulas, $$8(x-x_1)(x-x_2)(x-x_3) = 8x^3 + 4ax^2 + 2bx + a$$ gives us that $a = -8x_1x_2x_3 = -256 \Rightarrow \mathrm{(A)}$.

## See also

 2004 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 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

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