Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

(x−2)(x+3)(x−1)^2  ≥ 0

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

Commented by ZiYangLee last updated on 04/Sep/20

x≤−3 ∪ x≥2?

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

Commented by weltr last updated on 04/Sep/20

the answer must be   (−∞, −3] ∪ {1} ∪ [2, +∞)

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

Answered by Her_Majesty last updated on 04/Sep/20

(x−1)^2 ≥0∀x∈R but (x−1)^2 =0; x=1  (x−2)(x+3)≥0  ⇒ x≤−3 ∨ x≥2 ∨x=1

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com