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Question Number 151622 by otchereabdullai@gmail.com last updated on 22/Aug/21

$$\mathrm{Find}\mathrm{the}\mathrm{inequality}\mathrm{with}\mathrm{integer}$$$$\mathrm{coefficient}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{solution}\mathrm{set}$$$$\mathrm{x}:\mathrm{x}<\frac{-2}{5}\mathrm{or}>\frac{2}{5}$$

Answered by Rasheed.Sindhi last updated on 22/Aug/21

$$\mathrm{x}<\frac{-2}{5}\mathrm{or}>\frac{2}{5}$$$$5\mathrm{x}<-2\mathrm{or}5\mathrm{x}>2$$$$5\mathrm{x}+2<0\mathrm{or}5\mathrm{x}-2>0$$$$(5\mathrm{x}+2)(5\mathrm{x}-2)<0$$$${25\mathrm{x}}^2-4<0$$

Commented by mr W last updated on 22/Aug/21

$$5\mathrm{x}+2<0\mathrm{or}5\mathrm{x}-2>0$$$$when5x+2<0,then5x-2<0$$$$or$$$$when5x-2>0,then5x+2>0$$$$inbothcases:$$$$(5\mathrm{x}+2)(5\mathrm{x}-2)>0not<0$$$$i.e.25x^2-4>0not<0$$

Commented by mr W last updated on 22/Aug/21

$$\Rightarrow \{\begin{array}{l}\mathrm{x}>-\frac{2}{5}\mathrm{\&}x>\frac{2}{5}\Rightarrow x>\frac{2}{5}\\ \mathrm{OR}\Rightarrow or\\ x<-\frac{2}{5}\mathrm{\&}x<\frac{2}{5}\Rightarrow x<-\frac{2}{5}\end{array}$$$$thisisexactlyx<-\frac{2}{5}orx>\frac{2}{5}$$$$whichisrequested.$$$$whyyouchangeditto$$$$x>-\frac{2}{5}orx<\frac{2}{4}?$$

Commented by otchereabdullai@gmail.com last updated on 22/Aug/21

$$\mathrm{ok}!\mathrm{i}\mathrm{now}\mathrm{see}\mathrm{how}\mathrm{to}\mathrm{aproach}\mathrm{such}$$$$\mathrm{questions}\mathrm{thanks}\mathrm{a}\mathrm{lot}\mathrm{sir}!\mathrm{Rasheed}$$$$\mathrm{and}\mathrm{Prof}\mathrm{W}$$

Commented by Rasheed.Sindhi last updated on 22/Aug/21

$$mr\mathrm{W}\mathit{s}\mathit{i}\mathit{r}$$$$(5\mathrm{x}+2)(5\mathrm{x}-2)>0$$$$\Rightarrow \{\begin{array}{l}5\mathrm{x}+2>0\mathrm{\&}5x-2>0\\ \mathrm{OR}\\ 5x+2<0\mathrm{\&}5x-2<0\end{array}$$$$\Rightarrow \{\begin{array}{l}\mathrm{x}>-\frac{2}{5}\mathrm{\&}x>\frac{2}{5}\\ \mathrm{OR}\\ x<-\frac{2}{5}\mathrm{\&}x<\frac{2}{5}\end{array}$$$$\nRightarrow x\check{<}-\frac{2}{5}orx<\frac{2}{5}$$$${Where}^{\prime }serrorsir?$$

Commented by Rasheed.Sindhi last updated on 22/Aug/21

$$\underset{\underset{\underset{\underset{\underset{\mathit{S}}{\mathit{K}}}{\mathit{N}}}{\mathbf{A}}}{\mathrm{H}}}{\mathit{T}}\mathit{A}\underset{\underset{\mathit{T}}{\mathit{O}}}{\mathit{L}}\underset{\underset{\underset{!}{\mathit{R}}}{\mathit{I}}}{\mathit{S}}Icorrectedmymistake.$$

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