题目

Problem

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

\begin{align*} \left|2x^2 + x - 3\right| > 3|1 - x| \end{align*}

(7)

Solutions based entirely on graphical or numerical methods are not acceptable.

解答

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先把左边因式分解：

\begin{align*} 2x^2+x-3=(2x+3)(x-1). \end{align*}

这样两边都有 $|x-1|$ 。但 $x=1$ 时两边都是 $0$ ，不满足严格大于，所以之后可以在 $x\neq1$ 的条件下除以正数 $|x-1|$ 。

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Factorise the expression inside the modulus:

\begin{align*} 2x^2+x-3=(2x+3)(x-1). \end{align*}

\begin{align*} \left|2x^2+x-3\right| =&\,|(2x+3)(x-1)|\\[4mm] =&\,|2x+3|\,|x-1|. \end{align*}

The inequality becomes

\begin{align*} |2x+3|\,|x-1|>3|1-x|. \end{align*}

Since

\begin{align*} |1-x|=|x-1|, \end{align*}

we have

\begin{align*} |2x+3|\,|x-1|>3|x-1|. \end{align*}

If $x=1$ , then both sides are $0$ , so $x=1$ is not a solution. For $x\neq1$ , $|x-1|>0$ , so we can divide both sides by $|x-1|$ :

\begin{align*} |2x+3|>3. \end{align*}

Therefore,

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

This gives

\begin{align*} x>0 \quad \text{or} \quad x<-3. \end{align*}

Remembering that $x=1$ is not allowed, the solution set is

\begin{align*} x<-3, \qquad 0<x<1, \qquad x>1. \end{align*}

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因为两边都是非负数，也可以平方。这样会得到一个 quartic inequality，再因式分解。这个方法较长，但完全符合 “Use algebra” 的要求。

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Since both sides are non-negative, square both sides:

\begin{align*} \left(2x^2+x-3\right)^2>9(1-x)^2. \end{align*}

Bring everything to one side:

\begin{align*} \left(2x^2+x-3\right)^2-9(1-x)^2>0. \end{align*}

This is a difference of squares:

\begin{align*} &\,\left(2x^2+x-3-3(1-x)\right)\\[2mm] &\,\hspace{2pt}\times \left(2x^2+x-3+3(1-x)\right)>0. \end{align*}

Simplify each factor:

\begin{align*} 2x^2+x-3-3(1-x) =&\,2x^2+4x-6\\[2mm] =&\,2(x+3)(x-1), \end{align*}

and

\begin{align*} 2x^2+x-3+3(1-x) =&\,2x^2-2x\\[2mm] =&\,2x(x-1). \end{align*}

Therefore,

\begin{align*} 4x(x+3)(x-1)^2>0. \end{align*}

Since $4>0$ and $(x-1)^2\geqslant0$ , with equality only at $x=1$ , the sign is controlled by $x(x+3)$ , except that $x=1$ is not a solution.

So we need

\begin{align*} x(x+3)>0 \quad \text{and} \quad x\neq1. \end{align*}

Thus

\begin{align*} x<-3, \qquad 0<x<1, \qquad x>1. \end{align*}