Square substitution preserves an excluded zero · source reference

Complete question

Without a graphing calculator solve $x\le15/(x+2)$. [3] Hence solve $1/x^2\ge(x^2+2)/15$. [2]

[2009/NJC/Prelim/I/1]

Printed working

The instruction says a graphic calculator is not to be used.

$$\begin{aligned} x&\le\frac{15}{x+2},\qquad x\ne-2\\ x-\frac{15}{x+2}&\le0\\ \frac{x^2+2x-15}{x+2}&\le0\\ \frac{(x+5)(x-3)}{x+2}&\le0\\ (x+2)(x+5)(x-3)&\le0,\qquad x\ne-2. \end{aligned}$$

Thus $x\le-5$ or $-2<x\le3$.

$$\frac1{x^2}\ge\frac{x^2+2}{15}\quad\Longrightarrow\quad x^2\le\frac{15}{x^2+2},$$

since $x^2,x^2+2>0$ and $x\ne0$.

Using the earlier result with $x$ replaced by $x^2$ gives $x^2\le-5$ or $-2<x^2\le3$. The first has no real solutions since $x^2>0$; the second becomes $0<x^2\le3$.

Therefore $-\sqrt3\le x<0$ or $0<x\le\sqrt3$. Alternatively, $-\sqrt3\le x\le\sqrt3$, $x\ne0$.

Teacher margin notes

Can sketch the graph of $y=x^2$.

Source diagrams

Provenance

Question page references: 2. Solution page references: 11, 12. 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.