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 $a = -8x_1x_2x_3 = -256 \Rightarrow \mathrm{(A)}$ .

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 12 Problems and Solutions

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2004 AMC 12B Problems/Problem 17

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