题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

The equation

\begin{align*} 4(p-2x)=\frac{12+15p}{x+p},\qquad x\ne -p, \end{align*}

where $p$ is a constant, has two distinct real roots.

(a) Show that

\begin{align*} 3p^2-10p-8>0. \end{align*}

(3)

(b) Hence, using algebra, find the range of possible values of $p$ .

(3)

解答

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先把方程化成关于 $x$ 的二次方程。因为有两个不同实根，所以判别式必须大于 $0$ 。题目要的是关于 $p$ 的不等式。

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Starting with

\begin{align*} 4(p-2x)=\frac{12+15p}{x+p}, \end{align*}

multiply by $x+p$ :

\begin{align*} 4(p-2x)(x+p)=&\,12+15p. \end{align*}

Expand the left-hand side:

\begin{align*} (p-2x)(x+p) =&\,px+p^2-2x^2-2px\\ =&\,p^2-px-2x^2. \end{align*}

\begin{align*} 4p^2-4px-8x^2=&\,12+15p\\ 8x^2+4px+12+15p-4p^2=&\,0. \end{align*}

For two distinct real roots,

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

Here

\begin{align*} a=8,\qquad b=4p,\qquad c=12+15p-4p^2. \end{align*}

Therefore

\begin{align*} (4p)^2-4(8)(12+15p-4p^2)&>0\\ 16p^2-32(12+15p-4p^2)&>0\\ 16p^2-384-480p+128p^2&>0\\ 144p^2-480p-384&>0\\ 3p^2-10p-8&>0. \end{align*}

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承接 (a)，解二次不等式。因为二次项系数为正，所以大于 $0$ 的范围在两个根的外侧。

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Solve

\begin{align*} 3p^2-10p-8&>0. \end{align*}

Factorise:

\begin{align*} 3p^2-10p-8 =&\,(p-4)(3p+2). \end{align*}

So the critical values are

\begin{align*} p=4 \quad\text{and}\quad p=-\frac23. \end{align*}

Since the quadratic opens upwards,

\begin{align*} (p-4)(3p+2)>0 \end{align*}

outside the two roots.

Therefore

\begin{align*} p<-\frac23 \quad\text{or}\quad p>4. \end{align*}