题目

Problem

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

Find the range of values of $k$ for which the quadratic equation

\begin{align*} kx^2+8x+2(k+7)=0 \end{align*}

has no real roots.

(6)

解答

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二次方程没有实根时，判别式必须小于 $0$ 。这里变量是 $x$ ，参数是 $k$ ，所以把 $a=k$ 、 $b=8$ 、 $c=2(k+7)$ 代入 $b^2-4ac<0$ ，再解关于 $k$ 的二次不等式。

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For no real roots,

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

Here

\begin{align*} a=k,\qquad b=8,\qquad c=2(k+7). \end{align*}

\begin{align*} 8^2-4(k)\{2(k+7)\}&<0\\ 64-8k(k+7)&<0\\ 64-8k^2-56k&<0. \end{align*}

Divide by $-8$ , remembering to reverse the inequality:

\begin{align*} k^2+7k-8&>0\\ (k+8)(k-1)&>0. \end{align*}

The critical values are $k=-8$ and $k=1$ .

Since the product is positive outside the two roots,

\begin{align*} k<-8\quad\text{or}\quad k>1. \end{align*}