AMC12 2007 A

AMC12 2007 A · Q21

AMC12 2007 A · Q21. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

The sum of the zeros, the product of the zeros, and the sum of the coefficients of the function $f(x)=ax^{2}+bx+c$ are equal. Their common value must also be which of the following?

函数 $f(x)=ax^{2}+bx+c$ 的零点的和、零点的乘积以及系数的和相等。它们的共同值还必须是下列哪一项？

(A) the coefficient of $x^2$ $x^2$ 的系数

(B) the coefficient of $x$ $x$ 的系数

(D) one of the x-intercepts of the graph of $y = f(x)$ $y = f(x)$ 图形的某个 $x$ 截距

(E) the mean of the x-intercepts of the graph of $y = f(x)$ $y = f(x)$ 图形的 $x$ 截距的平均值

Answer

Correct choice: (A)

正确答案：（A）

Solution

By Vieta's Formulas, the sum of the roots of a quadratic equation is $\frac {-b}a$, the product of the zeros is $\frac ca$, and the sum of the coefficients is $a + b + c$. Setting equal the first two tells us that $\frac {-b}{a} = \frac ca \Rightarrow b = -c$. Thus, $a + b + c = a + b - b = a$, so the common value is also equal to the coefficient of $x^2 \Longrightarrow \fbox{\textrm{A}}$. To disprove the others, note that: - $\mathrm{B}$: then $b = \frac {-b}a$, which is not necessarily true. - $\mathrm{C}$: the y-intercept is $c$, so $c = \frac ca$, not necessarily true. - $\mathrm{D}$: an x-intercept of the graph is a root of the polynomial, but this excludes the other root. - $\mathrm{E}$: the mean of the x-intercepts will be the sum of the roots of the quadratic divided by 2.

由韦达定理，二次方程的两根之和为 $\frac {-b}a$，两根之积为 $\frac ca$，而系数之和为 $a + b + c$。令前两者相等得 $\frac {-b}{a} = \frac ca \Rightarrow b = -c$。因此 $a + b + c = a + b - b = a$，所以共同值也等于 $x^2$ 的系数 $\Longrightarrow \fbox{\textrm{A}}$。为否定其他选项，注意： - $\mathrm{B}$：则需有 $b = \frac {-b}a$，这不一定成立。 - $\mathrm{C}$：$y$ 截距为 $c$，则需有 $c = \frac ca$，不一定成立。 - $\mathrm{D}$：图像的一个 $x$ 截距是多项式的一个根，但这排除了另一个根。 - $\mathrm{E}$：$x$ 截距的平均值为二次方程两根之和除以 2。

Topics