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 80405 by TawaTawa last updated on 02/Feb/20

$$\mathrm{Solve}:$$ $$\left(\mathrm{a}\right){(\mathrm{x}-3)}^2>-5$$ $$\left(\mathrm{b}\right){3\mathrm{x}}^2>-12$$

Commented byMJS last updated on 02/Feb/20

$$\mathrm{a}{(x-3)}^2⩾0>-5\Rightarrow \mathrm{true}\mathrm{for}\mathrm{all}x\in \mathbb{R}$$ $$\mathrm{b}3x^2⩾0>-12\Rightarrow \mathrm{true}\mathrm{for}\mathrm{all}x\in \mathbb{R}$$ $$\mathrm{strange}\mathrm{examples}...$$

Commented byTawaTawa last updated on 02/Feb/20

$$\mathrm{Sir},\mathrm{no}\mathrm{workings}??$$

Commented byMJS last updated on 02/Feb/20

$$\mathrm{no}\mathrm{workings}.$$ $$\mathrm{\forall }r\in \mathbb{R}:r^2⩾0$$ $$\mathrm{nothing}\mathrm{else}\mathrm{is}\mathrm{necessary}$$

Commented byTawaTawa last updated on 02/Feb/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented bysandy_delta last updated on 03/Feb/20

$$howaboutcomplexnumber?$$

Commented byjagoll last updated on 03/Feb/20

$$\left(a\right){(x-3)}^2+5>0$$ $$(x-3-i\sqrt{5})(x-3+i\sqrt{5})>0$$ $$x<3-i\sqrt{5}\vee x>3+i\sqrt{5}$$

Commented byjagoll last updated on 03/Feb/20

$$\left(b\right)3x^2+12>0$$ $$3(x^2+4)>0$$ $$3(x+2i)(x-2i)>0$$ $$x<-2i\vee x>2i$$

Commented byMJS last updated on 03/Feb/20

$$\mathrm{complex}\mathrm{numbers}\mathrm{cannot}\mathrm{be}\mathrm{sorted}\mathrm{using}$$ $$<\mathrm{or}>\mathrm{so}\mathrm{this}\mathrm{makes}\mathrm{no}\mathrm{sense}\mathrm{at}\mathrm{all}$$

Commented byjagoll last updated on 03/Feb/20

$$-2i<2isir?$$

Commented byMJS last updated on 03/Feb/20

$$-2\mathrm{i}<2\mathrm{i}\mid -2\mathrm{i}$$ $$-4\mathrm{i}<0\mid \mathbf{÷}(-4)$$ $$\mathrm{i}>0$$ $$\mathrm{we}\mathrm{are}\mathrm{allowed}\mathrm{to}\mathrm{multiply}\mathrm{both}\mathrm{sides}\mathrm{with}$$ $$\mathrm{a}\mathrm{number}>0\mathrm{without}\mathrm{changing}\mathrm{the}<>\mathrm{signs}$$ $$\mathrm{i}>0\mid \times \mathrm{i}$$ $${\mathrm{i}}^2>0$$ $$-1>0$$ $$\mathrm{wrong}$$ $$$$ $$\mathrm{i}<0???$$ $$\mathrm{we}\mathrm{must}\mathrm{change}\mathrm{the}<>\mathrm{signs}\mathrm{when}\mathrm{multiplying}$$ $$\mathrm{with}\mathrm{a}\mathrm{number}<0$$ $$\mathrm{i}<0\mid \times \mathrm{i}$$ $${\mathrm{i}}^2>0$$ $$-1>0$$ $$\mathrm{wrong}$$ $$$$ $$\Rightarrow \mathrm{i}isneither>0nor<0$$

Commented byjagoll last updated on 03/Feb/20

$$thankyoumister$$

Answered by sandy_delta last updated on 03/Feb/20

$$\left(b\right)3x^2>-12$$ $$x^2>-4$$ $$x^2-(-4)>0$$ $$(x+\sqrt{-4})(x-\sqrt{-4})>0$$ $$(x+2i)(x-2i)>0$$ $$x<-2i\vee x>2i$$

Commented byMJS last updated on 03/Feb/20

$$\mathrm{see}\mathrm{my}\mathrm{above}\mathrm{comment}\mathrm{and}/\mathrm{or}\mathrm{please}\mathrm{tell}$$ $$\mathrm{me}\mathrm{if}$$ $$-3\mathrm{i}+5\mathrm{is}\mathrm{smaller}\mathrm{or}\mathrm{greater}\mathrm{than}-2\mathrm{i}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com