AMC12 2025 B

AMC12 2025 B · Q8

AMC12 2025 B · Q8. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

There are integers $a$ and $b$ such that the polynomial $x^3 - 5x^2 + ax + b$ has $4+\sqrt{5}$ as a root. What is $a+b$?

存在整数 $a$ 和 $b$，使得多项式 $x^3 - 5x^2 + ax + b$ 以 $4+\sqrt{5}$ 为根。求 $a+b$。

(A) 13 13

(B) 17 17

(D) 30 30

(E) 68 68

Answer

Correct choice: (C)

正确答案：（C）

Solution

Because every coefficient of our polynomial $f(x)$ is an integer, the conjugate of $4+\sqrt{5}$, $4-\sqrt{5}$ is also a root. Let the third and final root be $r_3$. Thanks to Vieta's Formulas, we know that the sum of the roots is \[(r_3) + (4+\sqrt{5}) + (4 - \sqrt{5}) = 5\] \[r_3 = -3.\] We now calculate $f(1) = (1+3)(1-4-\sqrt{5})(1-4+\sqrt{5})$. $(1-4-\sqrt{3})(1-4+\sqrt{3}) = (-3-\sqrt{5})(-3+\sqrt{5}) = 9 - 5 = 4.$ So $f(1) = 4 \cdot 4 = 16.$ Hence $f(1) = 16 = 1-5 + a + b$. Therefore, $a + b = \boxed{20}$.

因为我们的多项式 $f(x)$ 的每个系数都是整数，其共轭根 $4-\sqrt{5}$ 也是根。设第三个根为 $r_3$。由 Vieta 公式，我们知道根的和为 \[(r_3) + (4+\sqrt{5}) + (4 - \sqrt{5}) = 5\] \[r_3 = -3.\] 我们计算 $f(1) = (1+3)(1-4-\sqrt{5})(1-4+\sqrt{5})$。 $(1-4-\sqrt{5})(1-4+\sqrt{5}) = (-3-\sqrt{5})(-3+\sqrt{5}) = 9 - 5 = 4$。所以 $f(1) = 4 \cdot 4 = 16$。因此 $f(1) = 16 = 1-5 + a + b$。故 $a + b = \boxed{20}$。

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