Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 192570 by mechanics last updated on 21/May/23

Answered by cortano12 last updated on 21/May/23

 ∣x^2 −4∣<5   (x^2 −9)(x^2 +1)<0   (x+3)(x−3)(x^2 +1)<0    ∴ −3 < x < 3

$$\:\mid\mathrm{x}^{\mathrm{2}} −\mathrm{4}\mid<\mathrm{5} \\ $$$$\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)<\mathrm{0} \\ $$$$\:\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)<\mathrm{0} \\ $$$$\:\:\therefore\:−\mathrm{3}\:<\:\mathrm{x}\:<\:\mathrm{3}\: \\ $$

Answered by Skabetix last updated on 21/May/23

∣x∣+∣x+2∣+∣2−x∣≤8⇒−(8/3)≤x≤(8/3)

$$\mid{x}\mid+\mid{x}+\mathrm{2}\mid+\mid\mathrm{2}−{x}\mid\leqslant\mathrm{8}\Rightarrow−\frac{\mathrm{8}}{\mathrm{3}}\leqslant{x}\leqslant\frac{\mathrm{8}}{\mathrm{3}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com