题目

Problem

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

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

\begin{align*} |x^2-9|<|1-2x| \end{align*}

(6)

解答

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绝对值相等的位置是临界点。分别解 $x^2-9=1-2x$ 和 $x^2-9=-(1-2x)$ ，再用数轴判断哪些区间满足左边距离更小。

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The critical values occur when

\begin{align*} x^2-9=1-2x \end{align*}

\begin{align*} x^2-9=-(1-2x) \end{align*}

First,

\begin{align*} x^2-9 ={}& 1-2x\\[3mm] x^2+2x-10 ={}& 0\\[3mm] x ={}& -1\pm\sqrt{11} \end{align*}

Next,

\begin{align*} x^2-9 ={}& -1+2x\\[3mm] x^2-2x-8 ={}& 0\\[3mm] (x-4)(x+2) ={}& 0 \end{align*}

\begin{align*} x=-2,\quad 4 \end{align*}

The critical values in increasing order are

\begin{align*} -1-\sqrt{11},\quad -2,\quad -1+\sqrt{11},\quad 4 \end{align*}

Testing the intervals between these values gives

\begin{align*} \boxed{ -1-\sqrt{11}<x<-2 \quad\text{or}\quad -1+\sqrt{11}<x<4 } \end{align*}

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因为两边都是非负数，可以平方。平方后得到四次方程；但重点仍然是把它因式分解，再从数轴判断不等式的符号。

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Since both sides are non-negative,

\begin{align*} |x^2-9|<|1-2x| \end{align*}

is equivalent to

\begin{align*} (x^2-9)^2<(1-2x)^2 \end{align*}

Bring all terms to one side:

\begin{align*} (x^2-9)^2-(1-2x)^2 &<0\\[3mm] \left(x^2-9-(1-2x)\right) \left(x^2-9+(1-2x)\right) &<0\\[3mm] (x^2+2x-10)(x^2-2x-8) &<0\\[3mm] (x+1-\sqrt{11})(x+1+\sqrt{11})(x-4)(x+2) &<0 \end{align*}

Using a sign diagram, the product is negative on

\begin{align*} \boxed{ -1-\sqrt{11}<x<-2 \quad\text{or}\quad -1+\sqrt{11}<x<4 } \end{align*}