题目

Problem

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

\begin{align*} \vert x^2 + x - 2\vert < \frac{1}{2}(x + 5) \end{align*}

(7)

解答

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绝对值不等式 $|A|<B$ 可以理解成 $A$ 必须同时小于 $B$ ，并且大于 $-B$ 。所以这里把它拆成两个不等式：

\begin{align*} x^2+x-2 < \frac12(x+5) \end{align*}

和

\begin{align*} -(x^2+x-2) < \frac12(x+5). \end{align*}

最后要取两个结果的交集。

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We need

\begin{align*} -\frac12(x+5)<x^2+x-2<\frac12(x+5). \end{align*}

First solve

\begin{align*} x^2+x-2<&\,\frac12(x+5)\\[4mm] 2x^2+2x-4<&\,x+5\\[4mm] 2x^2+x-9<&\,0. \end{align*}

The critical values are

\begin{align*} x=&\,\frac{-1\pm\sqrt{1+72}}{4}\\[4mm] =&\,\frac{-1\pm\sqrt{73}}{4}. \end{align*}

Since the quadratic opens upwards,

\begin{align*} \frac{-1-\sqrt{73}}{4}<x<\frac{-1+\sqrt{73}}{4}. \end{align*}

Now solve

\begin{align*} -(x^2+x-2)<&\,\frac12(x+5)\\[4mm] -2x^2-2x+4<&\,x+5\\[4mm] 2x^2+3x+1>&\,0\\[4mm] (2x+1)(x+1)>&\,0. \end{align*}

Therefore,

\begin{align*} x<-1 \quad \text{or} \quad x>-\frac12. \end{align*}

Taking the intersection of the two conditions gives

\begin{align*} \frac{-1-\sqrt{73}}{4}<x<-1 \quad \text{or} \quad -\frac12<x<\frac{-1+\sqrt{73}}{4}. \end{align*}