x-2-x-3-x-1-2-0-

Question Number 111623 by weltr last updated on 04/Sep/20

$$\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{0} \\ $$

Commented by ZiYangLee last updated on 04/Sep/20

$${x}\leqslant−\mathrm{3}\:\cup\:{x}\geqslant\mathrm{2}? \\ $$

Commented by weltr last updated on 04/Sep/20

$${the}\:{answer}\:{must}\:{be}\: \\ $$$$\left(−\infty,\:−\mathrm{3}\right]\:\cup\:\left\{\mathrm{1}\right\}\:\cup\:\left[\mathrm{2},\:+\infty\right) \\ $$

Answered by Her_Majesty last updated on 04/Sep/20

$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{0}\forall{x}\in\mathbb{R}\:{but}\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0};\:{x}=\mathrm{1} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\:{x}\leqslant−\mathrm{3}\:\vee\:{x}\geqslant\mathrm{2}\:\vee{x}=\mathrm{1} \\ $$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com