题目

Problem

Use algebra to obtain the set of values of $x$ for which

\begin{align*} \left| \frac{x^2+3x+10}{x+2} \right| <\,& 7 - x\\[2mm] \end{align*}

(9)

解答

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这是含模的不等式。因为左边是绝对值，所以左边一定非负；因此若不等式成立，右边也必须为正，即 $7-x>0$ 。

处理绝对值时可以找边界点：

\begin{align*} \frac{x^2+3x+10}{x+2}=7-x \end{align*}

和

\begin{align*} -\frac{x^2+3x+10}{x+2}=7-x. \end{align*}

再用这些临界值分区间判断。

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We need

\begin{align*} \left|\frac{x^2+3x+10}{x+2}\right|<7-x, \end{align*}

where

\begin{align*} x\neq -2. \end{align*}

The boundary values occur when

\begin{align*} \frac{x^2+3x+10}{x+2}=7-x \end{align*}

\begin{align*} -\frac{x^2+3x+10}{x+2}=7-x. \end{align*}

First solve

\begin{align*} \frac{x^2+3x+10}{x+2}=7-x. \end{align*}

Multiplying by $x+2$ gives

\begin{align*} x^2+3x+10=&\,(7-x)(x+2)\\[2mm] =&\,7x+14-x^2-2x\\[2mm] =&\,5x+14. \end{align*}

\begin{align*} 2x^2-2x-4=&\,0\\[2mm] x^2-x-2=&\,0\\[2mm] (x-2)(x+1)=&\,0. \end{align*}

Hence

\begin{align*} x=2,\qquad x=-1. \end{align*}

Now solve

\begin{align*} -\frac{x^2+3x+10}{x+2}=7-x. \end{align*}

Multiplying by $x+2$ gives

\begin{align*} -(x^2+3x+10)=&\,(7-x)(x+2)\\[2mm] -x^2-3x-10=&\,5x+14-x^2\\[2mm] -3x-10=&\,5x+14\\[2mm] -8x=&\,24\\[2mm] x=&\,-3. \end{align*}

The critical values are therefore

\begin{align*} -3,\quad -2,\quad -1,\quad 2. \end{align*}

Testing the intervals, remembering that $x=-2$ is not allowed, gives

\begin{align*} x<-3 \quad\text{or}\quad -1<x<2. \end{align*}

Therefore the solution set is

\begin{align*} \boxed{x<-3\quad\text{or}\quad -1<x<2}. \end{align*}