Question Number 33468 by NECx last updated on 17/Apr/18
$${The}\:{set}\:{of}\:{integers}\:{that}\:{satisfies} \\ $$$$\mathrm{5}>\mid{n}−\mathrm{2}\mid\geqslant\mid{n}+\mathrm{1}\mid\:{is} \\ $$
Answered by MJS last updated on 17/Apr/18
![25>(n−2)^2 n^2 −4n−21<0 n=2±(√(25))=2±5 ⇒ n∈]−3;7[ (n−2)^2 ≥(n+1)^2 6n≤3 n≤(1/2) ⇒ n∈]−∞;0] ]−3;7[ ∩ ]−∞;0]=]−3;0] ⇒ ⇒ n∈{−2; −1; 0}](https://www.tinkutara.com/question/Q33469.png)
$$\mathrm{25}>\left({n}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} −\mathrm{4}{n}−\mathrm{21}<\mathrm{0} \\ $$$$\left.{n}=\mathrm{2}\pm\sqrt{\mathrm{25}}=\mathrm{2}\pm\mathrm{5}\:\Rightarrow\:{n}\in\right]−\mathrm{3};\mathrm{7}\left[\right. \\ $$$$\left({n}−\mathrm{2}\right)^{\mathrm{2}} \geqslant\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{6}{n}\leqslant\mathrm{3} \\ $$$$\left.{n}\left.\leqslant\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{n}\in\right]−\infty;\mathrm{0}\right] \\ $$$$ \\ $$$$\left.\right]\left.−\left.\mathrm{3}\left.;\mathrm{7}\left[\:\cap\:\right]−\infty;\mathrm{0}\right]=\right]−\mathrm{3};\mathrm{0}\right]\:\Rightarrow \\ $$$$\Rightarrow\:{n}\in\left\{−\mathrm{2};\:−\mathrm{1};\:\mathrm{0}\right\} \\ $$
Commented by NECx last updated on 17/Apr/18
$${wow}…..\:{lots}\:{of}\:{thanks}\:{sir}.\:{I}'{m}\circledast \\ $$$${most}\:{grateful}. \\ $$