题目

Problem

(a) Find, using algebra, all real solutions of

\begin{align*} 2x^3+3x^2-35x=0. \end{align*}

(3)

(b) Hence find all real solutions of

\begin{align*} 2(y-5)^6+3(y-5)^4-35(y-5)^2=0. \end{align*}

(4)

解答

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先提出公因式 $x$ ，再解剩下的二次方程。

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\begin{align*} 2x^3+3x^2-35x=&\,0\\ x(2x^2+3x-35)=&\,0\\ x(2x-7)(x+5)=&\,0. \end{align*}

Therefore

\begin{align*} x=0,\quad x=\frac72,\quad x=-5. \end{align*}

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令 $x=(y-5)^2$ ，题目就变成上一问的形式。因为 $(y-5)^2\geq0$ ，所以只使用上一问中非负的 $x$ 值。

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Let

\begin{align*} x=(y-5)^2. \end{align*}

Then the equation becomes

\begin{align*} 2x^3+3x^2-35x=0. \end{align*}

From part (a),

\begin{align*} x=0,\quad \frac72,\quad -5. \end{align*}

Since $x=(y-5)^2\geq0$ , reject $x=-5$ .

If $x=0$ ,

\begin{align*} (y-5)^2=&\,0\\ y=&\,5. \end{align*}

If $x=\frac72$ ,

\begin{align*} (y-5)^2=&\,\frac72\\ y-5=&\,\pm\sqrt{\frac72}\\ y=&\,5\pm\sqrt{\frac72}. \end{align*}

Therefore

\begin{align*} y=5,\qquad y=5+\sqrt{\frac72},\qquad y=5-\sqrt{\frac72}. \end{align*}