题目

Problem

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

Solve the equation

\begin{align*} 2y^2+1=\frac{15}{y^2} \end{align*}

(5)

解答

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先乘以 $y^2$ 消去分母，得到关于 $y^2$ 的二次方程。令 $u=y^2$ 会让结构更清楚。

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Since $y\neq0$ , multiply both sides by $y^2$ :

\begin{align*} 2y^2+1=&\,\frac{15}{y^2}\\ 2y^4+y^2=&\,15\\ 2y^4+y^2-15=&\,0. \end{align*}

Let

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

Then

\begin{align*} 2u^2+u-15=&\,0\\ (2u-5)(u+3)=&\,0. \end{align*}

\begin{align*} u=\frac52 \quad\text{or}\quad u=-3. \end{align*}

But $u=y^2\geqslant0$ , so $u=-3$ gives no real solution.

Hence

\begin{align*} y^2=&\,\frac52\\ y=&\,\pm\sqrt{\frac52}\\ =&\,\pm\frac{\sqrt{10}}{2}. \end{align*}

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也可以先把所有项移到一边，再直接把含 $y$ 和 $\frac1y$ 的因式拆出来。这个方法比较短，但要注意最后仍然是在解 $y^2$ 。

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Start with

\begin{align*} 2y^2+1-\frac{15}{y^2}=0. \end{align*}

This can be factorised as

\begin{align*} \left(2y-\frac5y\right) \left(y+\frac3y\right)=&\,0. \end{align*}

\begin{align*} 2y-\frac5y=&\,0 \quad\text{or}\quad y+\frac3y=0. \end{align*}

From the first equation,

\begin{align*} 2y-\frac5y=&\,0\\ 2y^2-5=&\,0\\ y^2=&\,\frac52\\ y=&\,\pm\frac{\sqrt{10}}{2}. \end{align*}

From the second equation,

\begin{align*} y+\frac3y=&\,0\\ y^2+3=&\,0, \end{align*}

which gives no real solution.

Therefore

\begin{align*} y=\pm\frac{\sqrt{10}}{2}. \end{align*}