题目

Problem

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

The equation

\begin{align*} k(3x^2+8x+9)=2-6x \end{align*}

where $k$ is a real constant, has no real roots.

Find the range of possible values for $k$ .

(7)

解答

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先把方程整理成关于 $x$ 的二次方程。没有实根表示判别式小于 $0$ ，然后解关于 $k$ 的二次不等式。

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Rearrange the equation:

\begin{align*} k(3x^2+8x+9)=&\,2-6x\\ 3kx^2+8kx+9k-2+6x=&\,0\\ 3kx^2+(8k+6)x+9k-2=&\,0. \end{align*}

For no real roots,

\begin{align*} b^2-4ac<0. \end{align*}

Here

\begin{align*} a=3k,\qquad b=8k+6,\qquad c=9k-2. \end{align*}

\begin{align*} (8k+6)^2-4(3k)(9k-2)&<0\\ 64k^2+96k+36-108k^2+24k&<0\\ -44k^2+120k+36&<0. \end{align*}

Divide by $-4$ :

\begin{align*} 11k^2-30k-9&>0\\ (11k+3)(k-3)&>0. \end{align*}

The critical values are

\begin{align*} k=-\frac{3}{11},\qquad k=3. \end{align*}

Since the quadratic is positive outside the roots,

\begin{align*} k<-\frac{3}{11} \quad\text{or}\quad k>3. \end{align*}