Replace the input by its magnitude · source reference

Complete question

(a) Solve $(x^2-6)/(2x)>1$ exactly. [3] Hence solve $(x^2-6)/(2|x|)\ge1$. [2] (b) Prove algebraically that $x^2-4x+9>0$ for all real x. Hence solve $(x^3+2x^2+x+14)/(x^2+5)>x+1$.

[2008/RJC/J1/T3 CT/9]

Printed working

Part (a)

$$\begin{aligned} \frac{x^2-6}{2x}&>1,\qquad x\ne0\\ \frac{x^2-2x-6}{2x}&>0\\ \frac{(x-1)^2-7}{2x}&>0\\ 2x\bigl(x-(1+\sqrt7)\bigr)\bigl(x-(1-\sqrt7)\bigr)&>0. \end{aligned}$$

Thus $1-\sqrt7<x<0$ or $x>1+\sqrt7$.

Replace $x$ with $|x|$ to get $\dfrac{|x|^2-6}{2|x|}>1$.

The non-strict version gives

$$1-\sqrt7\le|x|<0\quad\text{or}\quad|x|\ge1+\sqrt7.$$

The first interval has no solution since $|x|>0$ on the domain. The second gives $x\le-1-\sqrt7$ or $x\ge1+\sqrt7$.

The printed solution set is

$$(-\infty,-1-\sqrt7]\cup[1+\sqrt7,\infty),$$

or alternatively $\{x\in\mathbb R:x\le-1-\sqrt7\text{ or }x\ge1+\sqrt7\}$.

Part (b)

$$x^2-4x+9=(x-2)^2+5.$$

Since $(x-2)^2\ge0$ for all real values of $x$, $x^2-4x+9=(x-2)^2+5>0$ for all real values of $x$. (Shown.)

The inequality labelled (*) is

$$\frac{x^3+2x^2+x+14}{x^2+5}>x+1.$$

Because $x^2+5>0$ for all $x\in\mathbb R$,

$$\begin{aligned} x^3+2x^2+x+14&>(x+1)(x^2+5)\\ x^3+2x^2+x+14&>x^3+x^2+5x+5\\ x^2-4x+9&>0. \end{aligned}$$

Thus the solution is $x\in\mathbb R$.

Teacher margin notes

As $x^2-2x-6$ cannot be easily factorised: (1) use the quadratic formula to obtain the roots of $x^2-2x-6=0$, which gives the x-intercepts of the graph; OR (2) complete the square of $x^2-2x-6$ to obtain the factors.

For $(x^2-6)/(2x)\ge1$, the solution is $1-\sqrt7\le x<0$ or $x\ge1+\sqrt7$, as the numerator can be zero in this case, unlike the first case.

Since $x^2+5>0$ for all $x\in\mathbb R$, there is no sign change when both sides of inequality (*) are multiplied by $x^2+5$.

Source diagrams

Provenance

Question page references: 1. Solution page references: 3. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.