题目

Problem

In this question you must show detailed reasoning.

Solutions relying on calculator technology are not acceptable.

\begin{align*} f(x)=x^2(2x+1)-15x \end{align*}

(a) Solve

\begin{align*} f(x)=0. \end{align*}

(4)

(b) Hence solve

\begin{align*} y^{\frac43}\left(2y^{\frac23}+1\right)-15y^{\frac23}=0, \qquad y>0, \end{align*}

giving your answer in simplified surd form.

(2)

解答

展开

先展开并提取公因式 $x$ ，剩下的是一个二次式。题目要求展示推理，所以不要只写答案。

展开

Since

\begin{align*} f(x)=x^2(2x+1)-15x, \end{align*}

we solve

\begin{align*} x^2(2x+1)-15x=&\,0\\ 2x^3+x^2-15x=&\,0\\ x(2x^2+x-15)=&\,0. \end{align*}

Factorise the quadratic:

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

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

Therefore

\begin{align*} x=0, \qquad x=\frac52, \qquad x=-3. \end{align*}

展开

注意到 $y^{4/3}=(y^{2/3})^2$ ，所以这一题其实是在把上一小题中的 $x$ 换成 $y^{2/3}$ 。因为 $y>0$ ，所以 $y^{2/3}>0$ ，只能使用上一小题的正根。

展开

Let

\begin{align*} x=y^{\frac23}. \end{align*}

Then

\begin{align*} y^{\frac43}\left(2y^{\frac23}+1\right) -15y^{\frac23}=0 \end{align*}

corresponds to

\begin{align*} x^2(2x+1)-15x=0. \end{align*}

From part (a), the positive solution is

\begin{align*} x=\frac52. \end{align*}

Hence

\begin{align*} y^{\frac23}=&\,\frac52. \end{align*}

Raise both sides to the power $\frac32$ :

\begin{align*} y =&\,\left(\frac52\right)^{\frac32}\\ =&\,\sqrt{\left(\frac52\right)^3}\\ =&\,\sqrt{\frac{125}{8}}\\ =&\,\frac{5\sqrt{10}}{4}. \end{align*}