题目

Problem

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

(i) Using the laws of indices, solve

\begin{align*} 3^{2p-5}=\frac19\sqrt{27} \end{align*}

(3)

(ii) Find the real roots of the equation

\begin{align*} x^2+1=\frac{36}{x^2-4} \end{align*}

(4)

解答

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把右边全部写成 $3$ 的幂。两边底数相同后，就可以比较指数。

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\begin{align*} \frac19\sqrt{27} =&\,3^{-2}\cdot(3^3)^{1/2}\\ =&\,3^{-2}\cdot3^{3/2}\\ =&\,3^{-1/2}. \end{align*}

\begin{align*} 3^{2p-5}=&\,3^{-1/2}. \end{align*}

Equating the indices gives

\begin{align*} 2p-5=&\,-\frac12\\ 2p=&\,\frac92\\ p=&\,\frac94. \end{align*}

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方程只含 $x^2$ ，所以先交叉相乘，整理成关于 $x^2$ 的二次方程。解出 $x^2$ 后，再回到 $x$ 。

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Since $x^2-4\neq0$ , multiply both sides by $x^2-4$ :

\begin{align*} (x^2+1)(x^2-4)=&\,36\\ x^4-4x^2+x^2-4=&\,36\\ x^4-3x^2-40=&\,0. \end{align*}

Let

\begin{align*} u=x^2. \end{align*}

Then

\begin{align*} u^2-3u-40=&\,0\\ (u-8)(u+5)=&\,0. \end{align*}

\begin{align*} u=8 \quad\text{or}\quad u=-5. \end{align*}

Since $u=x^2\geqslant0$ , reject $u=-5$ .

Thus

\begin{align*} x^2=&\,8\\ x=&\,\pm2\sqrt2. \end{align*}