Question Number 215820 by Jubr last updated on 18/Jan/25 Commented by A5T last updated on 18/Jan/25 $$\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{generally}\:\mathrm{true},\:\mathrm{it}\:\mathrm{fails}\:\mathrm{when}\:\mathrm{c}=\mathrm{0}. \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 215811 by MATHEMATICSAM last updated on 18/Jan/25 $$\mathrm{If}\:\alpha,\:\beta,\:\gamma,\:\delta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{4}} \:+\:{x}^{\mathrm{3}} \:+\:{x}^{\mathrm{2}} \:+\:{x}\:+\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{find} \\ $$$$\alpha^{\mathrm{2021}} \:+\:\beta^{\mathrm{2021}} \:+\:\gamma^{\mathrm{2021}} \:+\:\delta^{\mathrm{2021}} \:. \\ $$ Answered by…
Question Number 215730 by MathematicalUser2357 last updated on 16/Jan/25 $$\boldsymbol{\mathrm{Determine}}\:\boldsymbol{{a}},\:\boldsymbol{{b}},\:\boldsymbol{{c}}\:\left[\mathrm{Lazy}\:\mathrm{problem}\right] \\ $$$$\mathrm{J181}-\mathrm{2}.\:{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{15}{x}−\mathrm{7}=\left({x}+{a}\right)^{\mathrm{3}} +{bx}+{c} \\ $$$$\mathrm{J182}-\left(\mathrm{1}\right)\:{x}^{\mathrm{3}} +{ax}+\mathrm{2}=\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{bx}+{c}\right) \\ $$ Answered by A5T last…
Question Number 215708 by essaad last updated on 15/Jan/25 Answered by A5T last updated on 15/Jan/25 $$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\Rightarrow\mathrm{n}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} =\mathrm{0}\Rightarrow\left(\mathrm{n}+\mathrm{y}\right)\left(\mathrm{n}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{ny}\right)=\mathrm{0}…\left(\mathrm{iii}\right) \\ $$$$\left(\mathrm{i}\right)−\left(\mathrm{ii}\right)\Rightarrow\mathrm{4n}^{\mathrm{3}} −\mathrm{4y}^{\mathrm{3}} −\mathrm{8n}+\mathrm{8y}=\mathrm{0}…
Question Number 215656 by MathematicalUser2357 last updated on 13/Jan/25 $$\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} \\ $$$$\mathrm{x}=\mathrm{2}\:\:,\:\mathrm{3x}^{\mathrm{2}}…
Question Number 215659 by Ari last updated on 13/Jan/25 Answered by Rasheed.Sindhi last updated on 13/Jan/25 $$\mathrm{2}\left(\overline {{abc}}\right)=\overline {{ab}}+\overline {{ba}}+\overline {{bc}}+\overline {{cb}}+\overline {{ac}}+\overline {{ca}} \\…
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…
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 $$\sqrt{\mathrm{y}}\:+\sqrt{\mathrm{x}}\:=\mathrm{5}\:;\:\sqrt{\mathrm{x}}\:\sqrt{\mathrm{y}}\:=\mathrm{8} \\…
Question Number 215525 by zetamaths last updated on 09/Jan/25 $$\Phi\::\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R}^{\mathrm{2}} \: \\ $$$$\:\:\:\left({x}.;{y}\right)\mid\rightarrow\left(\mathrm{2}{x}+\mathrm{3}{y}:\mathrm{3}{y}\right) \\ $$$${find} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\Phi\in\mathscr{L}\left(\mathbb{R}^{\mathrm{2}} \right)…
Question Number 215469 by Abdullahrussell last updated on 08/Jan/25 Answered by alephnull last updated on 08/Jan/25 $$\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) \\ $$$$=\left(\frac{\mathrm{27}}{\mathrm{108}}+\frac{\mathrm{4}}{\mathrm{108}}\right) \\ $$$${the}\:{root}\:{is}\:\frac{\mathrm{31}}{\mathrm{108}} \\ $$$$…