题目

Problem

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

The curve $C_1$ has equation

\begin{align*} y=(x+5)(3x+2)(2x-5). \end{align*}

The curve $C_2$ has equation

\begin{align*} y=3x^2-33x-50. \end{align*}

Use algebra to find the $x$ coordinates of the points of intersection of $C_1$ and $C_2$ .

(5)

解答

展开

交点处两条曲线的 $y$ 值相等。先把 $C_1$ 展开，再与 $C_2$ 相等并整理。由于两边常数项相同，整理后会出现一个 $x$ 因子。

展开

First expand $C_1$ :

\begin{align*} (x+5)(3x+2)=&\,3x^2+17x+10. \end{align*}

\begin{align*} (x+5)(3x+2)(2x-5) =&\,(3x^2+17x+10)(2x-5)\\ =&\,6x^3-15x^2+34x^2-85x+20x-50\\ =&\,6x^3+19x^2-65x-50. \end{align*}

At the points of intersection,

\begin{align*} 6x^3+19x^2-65x-50=&\,3x^2-33x-50. \end{align*}

Bring all terms to one side:

\begin{align*} 6x^3+16x^2-32x=&\,0\\ 2x(3x^2+8x-16)=&\,0\\ 2x(3x-4)(x+4)=&\,0. \end{align*}

Therefore

\begin{align*} x=0,\qquad x=\frac43,\qquad x=-4. \end{align*}