AMC12 2004 B

AMC12 2004 B · Q17

AMC12 2004 B · Q17. It mainly tests Vieta / quadratic relationships (basic), Logarithms (rare).

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 $a$?

对于某些实数 $a$ 和 $b$，方程 \[8x^3 + 4ax^2 + 2bx + a = 0\] 有三个不同的正根。如果这些根的以 $2$ 为底的对数之和为 $5$，那么 $a$ 的值是多少？

(A) -256 -256

(B) -64 -64

(D) 64 64

(E) 256 256

Answer

Correct choice: (A)

正确答案：（A）

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

设三个根为 $x_1,x_2,x_3$。 \[\log_2 x_1 + \log_2 x_2 + \log_2 x_3 = \log_2 x_1x_2x_3= 5 \Longrightarrow x_1x_2x_3 = 32\] 由韦达定理， \[8(x-x_1)(x-x_2)(x-x_3) = 8x^3 + 4ax^2 + 2bx + a\] 得到 $a = -8x_1x_2x_3 = -256 \Rightarrow \mathrm{(A)}$。

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