AMC10 2008 B

AMC10 2008 B · Q9

AMC10 2008 B · Q9. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

A quadratic equation $ax^2 - 2ax + b = 0$ has two real solutions. What is the average of the solutions?

二次方程 $ax^2 - 2ax + b = 0$ 有两个实根。两个根的平均值是多少？

(A) 1 1

(B) 2 2

(D) \frac{2b}{a} \frac{2b}{a}

(E) \sqrt{2a - b} \sqrt{2a - b}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): The quadratic formula implies that the two solutions are $x_1=\dfrac{2a+\sqrt{4a^2-4ab}}{2a}$ and $x_2=\dfrac{2a-\sqrt{4a^2-4ab}}{2a}$, so the average is $\dfrac{1}{2}(x_1+x_2)=\dfrac{1}{2}\left(\dfrac{2a}{2a}+\dfrac{2a}{2a}\right)=1.$

答案（A）：二次公式表明两个解为 $x_1=\dfrac{2a+\sqrt{4a^2-4ab}}{2a}$ 和 $x_2=\dfrac{2a-\sqrt{4a^2-4ab}}{2a}$，因此它们的平均值是 $\dfrac{1}{2}(x_1+x_2)=\dfrac{1}{2}\left(\dfrac{2a}{2a}+\dfrac{2a}{2a}\right)=1.$

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