题目

Problem

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

\begin{align*} (k-1)x^6+4x^3+(k-4)=0 \end{align*}

where $k$ is a constant.

(a) Find the exact solutions to the given equation for $k=4.5$ .

(3)

(b) Find the set of possible values of $k$ for which the given equation has no real roots.

(4)

解答

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代入 $k=4.5$ 后，方程只含 $x^6$ 和 $x^3$ 。令 $u=x^3$ ，就能化成二次方程。

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When $k=4.5$ ,

\begin{align*} (4.5-1)x^6+4x^3+(4.5-4)=&\,0\\ 3.5x^6+4x^3+0.5=&\,0. \end{align*}

Multiply by $2$ :

\begin{align*} 7x^6+8x^3+1=0. \end{align*}

Let

\begin{align*} u=x^3. \end{align*}

Then

\begin{align*} 7u^2+8u+1=&\,0\\ (7u+1)(u+1)=&\,0. \end{align*}

\begin{align*} u=-\frac17 \quad\text{or}\quad u=-1. \end{align*}

Since $u=x^3$ ,

\begin{align*} x^3=&\,-\frac17 \quad\text{or}\quad x^3=-1. \end{align*}

Therefore the exact real solutions are

\begin{align*} x=-\sqrt[3]{\frac17} \quad\text{or}\quad x=-1. \end{align*}

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把方程看成关于 $u=x^3$ 的二次方程：

\begin{align*} (k-1)u^2+4u+(k-4)=0. \end{align*}

由于每一个实数 $u$ 都对应一个实数 $x=\sqrt[3]{u}$ ，所以原方程没有实根等价于这个二次方程没有实根。因此判别式要小于 $0$ 。

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Let

\begin{align*} u=x^3. \end{align*}

Then the equation becomes

\begin{align*} (k-1)u^2+4u+(k-4)=0. \end{align*}

For no real roots, its discriminant must be negative:

\begin{align*} 4^2-4(k-1)(k-4)&<0. \end{align*}

Simplify:

\begin{align*} 16-4(k-1)(k-4)&<0\\ 16-4(k^2-5k+4)&<0\\ 16-4k^2+20k-16&<0\\ -4k^2+20k&<0\\ 4k(5-k)&<0. \end{align*}

The critical values are

\begin{align*} k=0,\qquad k=5. \end{align*}

Since $4k(5-k)$ is negative outside these values,

\begin{align*} k<0\quad\text{or}\quad k>5. \end{align*}