题目

Problem

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

(a) Given

\begin{align*} y-x=4 \end{align*}

and

\begin{align*} 2x^2-xy=8 \end{align*}

show that

\begin{align*} x^2-4x-8=0 \end{align*}

(2)

(b) Hence solve the simultaneous equations

\begin{align*} y-x=4 \end{align*}

\begin{align*} 2x^2-xy=8 \end{align*}

writing the answers in the form $p+q\sqrt3$ , where $p$ and $q$ are integers to be found.

(4)

解答

展开

由 $y-x=4$ 得到 $y=x+4$ ，再代入第二个方程。因为题目要求 “show that”，中间展开步骤要写清楚。

展开

From

\begin{align*} y-x=4, \end{align*}

we have

\begin{align*} y=x+4 \end{align*}

Substitute this into $2x^2-xy=8$ :

\begin{align*} 2x^2-x(x+4) =&\,8\\[3mm] 2x^2-x^2-4x =&\,8\\[3mm] x^2-4x-8 =&\,0 \end{align*}

as required.

展开

用 (a) 得到的二次方程。这个方程不能整数因式分解，所以可以配方，得到含 $\sqrt3$ 的精确解。

展开

From part (a),

\begin{align*} x^2-4x-8=0 \end{align*}

Complete the square:

\begin{align*} x^2-4x-8 =&\,0\\[3mm] (x-2)^2-12 =&\,0\\[3mm] (x-2)^2 =&\,12\\[3mm] x-2 =&\,\pm2\sqrt3 \end{align*}

\begin{align*} x=2\pm2\sqrt3 \end{align*}

Since $y=x+4$ ,

\begin{align*} y=6\pm2\sqrt3 \end{align*}

Therefore the solutions are

\begin{align*} x=2+2\sqrt3,\quad y=6+2\sqrt3 \end{align*}

and

\begin{align*} x=2-2\sqrt3,\quad y=6-2\sqrt3 \end{align*}