Strict positivity requires a strict discriminant bound · source reference

Complete question

Show $x^2-x+1>0$ for every real x. Hence find all a such that $(x^2+ax-2)/(x^2-x+1)<2$ for every real x. [6]

[2008/TPJC/Prelim/I/4]

Printed working

$$x^2-x+1=\left(x-\frac12\right)^2+\frac34>0\quad\text{for all }x\in\mathbb R.\quad\text{(Shown.)}$$

For the inequality to hold for all real $x$,

$$\begin{aligned} \frac{x^2+ax-2}{x^2-x+1}&<2\\ \frac{x^2+ax-2-2(x^2-x+1)}{x^2-x+1}&<0\\ \frac{-x^2+(2+a)x-4}{x^2-x+1}&<0. \end{aligned}$$

Since $x^2-x+1>0$ for all real $x$, $-x^2+(2+a)x-4<0$ for all real $x$.

The printed curve lies entirely below the x-axis. Discriminant $<0$ (note: coefficient of $x^2=-1<0$):

$$\begin{aligned} (2+a)^2-4(-1)(-4)&<0\\ (2+a)^2-4^2&<0\\ (2+a+4)(2+a-4)&<0\\ (a+6)(a-2)&<0\\ -6&<a<2. \end{aligned}$$

Teacher margin notes

Show by completing the square.

Make one side of the inequality zero.

Since $-x^2+(2+a)x-4<0$ is “for all real x”, the quadratic curve must lie completely below the x-axis. To do so, the curve must have a maximum point. Thus two conditions must be satisfied: (1) discriminant $<0$ — the curve does not cut the x-axis, so $-x^2+(2+a)x-4=0$ has no real roots; (2) coefficient of $x^2<0$ — the curve has a maximum point.

Source diagrams

Provenance

Question page references: 2. Solution page references: 8. 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.