题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

(a) Fully factorise

\begin{align*} 9x^3-10x^2+x. \end{align*}

(2)

(b) Hence solve

\begin{align*} 9\times27^y-10\times9^y+3^y=0. \end{align*}

(3)

解答

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先提出公因式 $x$ ，剩下的是二次式 $9x^2-10x+1$ ，再因式分解。

展开

\begin{align*} 9x^3-10x^2+x =&\,x(9x^2-10x+1)\\ =&\,x(9x-1)(x-1). \end{align*}

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把 $27^y$ 和 $9^y$ 都写成 $3^y$ 的幂：

\begin{align*} 27^y=(3^3)^y=(3^y)^3,\qquad 9^y=(3^2)^y=(3^y)^2. \end{align*}

令 $X=3^y$ ，方程就变成 (a) 中的三次式。

展开

Let

\begin{align*} X=3^y. \end{align*}

Then

\begin{align*} 27^y=(3^y)^3=X^3,\qquad 9^y=(3^y)^2=X^2. \end{align*}

So the equation becomes

\begin{align*} 9X^3-10X^2+X=&\,0\\ X(9X-1)(X-1)=&\,0. \end{align*}

Since $X=3^y>0$ , we cannot have $X=0$ .

Thus

\begin{align*} 9X-1=0 \quad\text{or}\quad X-1=0. \end{align*}

\begin{align*} X=\frac19 \quad\text{or}\quad X=1. \end{align*}

Return to $X=3^y$ :

\begin{align*} 3^y=&\,\frac19=3^{-2} \quad\text{or}\quad 3^y=1=3^0. \end{align*}

Therefore

\begin{align*} y=-2\quad\text{or}\quad y=0. \end{align*}