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Category: Algebra

Question-215708

Question Number 215708 by essaad last updated on 15/Jan/25 Answered by A5T last updated on 15/Jan/25 $$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\Rightarrow\mathrm{n}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} =\mathrm{0}\Rightarrow\left(\mathrm{n}+\mathrm{y}\right)\left(\mathrm{n}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{ny}\right)=\mathrm{0}…\left(\mathrm{iii}\right) \\ $$$$\left(\mathrm{i}\right)−\left(\mathrm{ii}\right)\Rightarrow\mathrm{4n}^{\mathrm{3}} −\mathrm{4y}^{\mathrm{3}} −\mathrm{8n}+\mathrm{8y}=\mathrm{0}…

3x-3-2x-2-12x-8-0-

Question Number 215656 by MathematicalUser2357 last updated on 13/Jan/25 $$\mathrm{3}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{8}=\mathrm{0} \\ $$ Answered by Wuji last updated on 13/Jan/25 $$\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{2}\:\:,\:\mathrm{3x}^{\mathrm{2}}…

x-4-x-3-8x-2-2x-4-0-x-1-x-2-x-2-2-Is-this-right-I-have-not-enough-time-to-edit-my-solution-

Question Number 215625 by MathematicalUser2357 last updated on 12/Jan/25 $${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}=\mathrm{0} \\ $$$${x}=\mathrm{1}\:\vee\:{x}=\mathrm{2}\:\vee\:{x}=\mathrm{2}\pm\sqrt{\mathrm{2}} \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{right}?\:\mathrm{I}\:\mathrm{have}\:\mathrm{not}\:\mathrm{enough}\:\mathrm{time}\:\mathrm{to}\:\mathrm{edit}\:\mathrm{my}\:\mathrm{solution} \\ $$ Commented by MathematicalUser2357 last updated on…

y-x-5-x-y-8-x-y-x-y-y-x-

Question Number 215559 by hardmath last updated on 10/Jan/25 $$\sqrt{\mathrm{y}}\:+\:\sqrt{\mathrm{x}}\:=\:\mathrm{5} \\ $$$$\sqrt{\mathrm{x}}\:\centerdot\:\sqrt{\mathrm{y}}\:=\:\mathrm{8} \\ $$$$\frac{\sqrt{\mathrm{x}}\:\mathrm{y}\:−\:\mathrm{x}\:\sqrt{\mathrm{y}}}{\mathrm{y}\:−\:\mathrm{x}}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 10/Jan/25 $$\sqrt{\mathrm{y}}\:+\sqrt{\mathrm{x}}\:=\mathrm{5}\:;\:\sqrt{\mathrm{x}}\:\sqrt{\mathrm{y}}\:=\mathrm{8} \\…

Question-215469

Question Number 215469 by Abdullahrussell last updated on 08/Jan/25 Answered by alephnull last updated on 08/Jan/25 $$\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\right)=\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}\right) \\ $$$$=\left(\frac{\mathrm{27}}{\mathrm{108}}+\frac{\mathrm{4}}{\mathrm{108}}\right) \\ $$$${the}\:{root}\:{is}\:\frac{\mathrm{31}}{\mathrm{108}} \\ $$$$…

2-2-8-A-B-Find-A-B-

Question Number 215439 by hardmath last updated on 06/Jan/25 $$\left(\:\mathrm{2}\:\:+\:\:\sqrt{\mathrm{2}}\:\right)^{\mathrm{8}} \:=\:\:\sqrt{\mathrm{A}}\:\:+\:\:\sqrt{\mathrm{B}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{A}−\mathrm{B}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 07/Jan/25 $${If}\:{z}\:{is}\:{a}\:{binomial}\:{surd}\:{and}\:{n}\:{is} \\ $$$${natural}\:…