AMC10 2013 B

AMC10 2013 B · Q19

AMC10 2013 B · Q19. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

The real numbers c, b, a form an arithmetic sequence with a ≥ b ≥ c ≥ 0. The quadratic ax$^2$ + bx + c has exactly one root. What is this root?

实数 $c,b,a$ 构成公差序列，且 $a\ge b\ge c\ge0$。二次多项式 $ax^2+bx+c$ 恰有一个根。此根是什么？

(A) −7 - 4$\sqrt{3}$ −7 - 4$\sqrt{3}$

(B) −2 - $\sqrt{3}$ −2 - $\sqrt{3}$

(D) −2 + $\sqrt{3}$ −2 + $\sqrt{3}$

(E) −7 + 4$\sqrt{3}$ −7 + 4$\sqrt{3}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let the common difference in the arithmetic sequence be $d$, so that $a=b+d$ and $c=b-d$. Because the quadratic has exactly one root, $b^2-4ac=0$. Substitution gives $b^2=4(b+d)(b-d)$, and therefore $3b^2=4d^2$. Because $b\ge 0$ and $d\ge 0$, it follows that $\sqrt{3}b=2d$. Thus the real root is $$ \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-b}{2a} = \frac{-b}{2(b+d)} = \frac{-b}{2\left(b+\frac{\sqrt{3}}{2}b\right)} = -2+\sqrt{3}. $$ Note that the quadratic equation $x^2+(4-2\sqrt{3})x+7-4\sqrt{3}$ satisfies the given conditions.

答案（D）：设等差数列的公差为 $d$，则 $a=b+d$ 且 $c=b-d$。因为该二次方程恰有一个根，所以 $b^2-4ac=0$。代入得 $b^2=4(b+d)(b-d)$，因此 $3b^2=4d^2$。由于 $b\ge 0$ 且 $d\ge 0$，可得 $\sqrt{3}b=2d$。因此实根为 $$ \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-b}{2a} = \frac{-b}{2(b+d)} = \frac{-b}{2\left(b+\frac{\sqrt{3}}{2}b\right)} = -2+\sqrt{3}. $$ 注意：二次方程 $x^2+(4-2\sqrt{3})x+7-4\sqrt{3}$ 满足给定条件。

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