题目

Problem

(a) Find, using algebra, all solutions of

\begin{align*} 20x^3-50x^2-30x=0. \end{align*}

(3)

(b) Hence find all real solutions of

\begin{align*} 20(y+3)^{3/2}-50(y+3)-30(y+3)^{1/2}=0. \end{align*}

(4)

解答

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先提出公因式 $10x$ ，剩下的二次式再因式分解。

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\begin{align*} 20x^3-50x^2-30x=&\,0\\ 10x(2x^2-5x-3)=&\,0\\ 10x(2x+1)(x-3)=&\,0. \end{align*}

Therefore

\begin{align*} x=0,\quad x=-\frac12,\quad x=3. \end{align*}

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把 $(y+3)^{1/2}$ 看作上一问里的 $x$ 。因为平方根不能是负数，所以 $x=-\frac12$ 对应不到实数 $y$ ，要舍去。

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Let

\begin{align*} x=(y+3)^{1/2}. \end{align*}

Then the equation becomes

\begin{align*} 20x^3-50x^2-30x=0. \end{align*}

From part (a),

\begin{align*} x=0,\quad -\frac12,\quad 3. \end{align*}

Since $x=(y+3)^{1/2}\geq0$ , reject $x=-\frac12$ .

If $x=0$ ,

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

If $x=3$ ,

\begin{align*} (y+3)^{1/2}=&\,3\\ y+3=&\,9\\ y=&\,6. \end{align*}

Therefore

\begin{align*} y=-3,\quad 6. \end{align*}