A repeated factor does not disappear from a strict inequality · source reference

Complete question

Show $x^2(x+2)^2-1=(x+1)^2(x+1+\sqrt2)(x+1-\sqrt2)$. [1] Hence solve $1/(x+2)^2>x^2$ without a calculator. [3] Deduce the inputs satisfying $1/(e^x+2)^2>e^{2x}$. [2]

[modified 2008/RJC/Prelim/II/1]

Printed working

Identity

$$\begin{aligned} x^2(x+2)^2-1&=(x^2+2x)^2-1^2\\ &=(x^2+2x+1)(x^2+2x-1)\\ &=(x+1)^2\bigl[(x+1)^2-(\sqrt2)^2\bigr]\\ &=(x+1)^2(x+1+\sqrt2)(x+1-\sqrt2).\quad\text{(Shown.)} \end{aligned}$$

Rational inequality — reduced quadratic

$$\begin{aligned} \frac1{(x+2)^2}&>x^2,\qquad x\ne-2\\ x^2(x+2)^2&<1\qquad\text{since }(x+2)^2>0\\ (x+1)^2(x+1+\sqrt2)(x+1-\sqrt2)&<0,\qquad x\ne-2\\ (x+1+\sqrt2)(x+1-\sqrt2)&<0\qquad\text{since }(x+1)^2>0,\quad x\ne-1,-2. \end{aligned}$$

Thus $-1-\sqrt2<x<-2$ or $-2<x<-1$ or $-1<x<-1+\sqrt2$.

Alternatively write $-1-\sqrt2<x<-1+\sqrt2$, $x\ne-2,-1$.

Alternative method — quartic sketch

Directly solve $(x+1)^2(x+1+\sqrt2)(x+1-\sqrt2)<0$, $x\ne-2$, using the printed quartic. It crosses at $-1\pm\sqrt2$ and touches at $-1$; the excluded input $-2$ is marked. It gives the same three intervals and the same compact answer above.

Exponential substitution

Replace $x$ by $e^x$:

$$-1-\sqrt2<e^x<-2\quad\text{or}\quad-2<e^x<-1\quad\text{or}\quad-1<e^x<-1+\sqrt2.$$

Since $e^x>0$, only $0<e^x<-1+\sqrt2$ remains, giving $x<\ln(\sqrt2-1)$.

Teacher margin notes

Rewrite as $a^2-b^2=(a-b)(a+b)$.

Always state the values that $x$ cannot take so as to exclude these values from the solution.

Sketch the graph of $y=e^x$ and, by looking at the portion of graph between $y=0$ and $y=\sqrt2-1$, determine the range of values of $x$.

Source diagrams

Provenance

Question page references: 2. Solution page references: 10, 11. 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.