Question Number 65480 by ajfour last updated on 30/Jul/19 $${x}^{\mathrm{4}} −\mathrm{15}{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{24}=\mathrm{0}\:\:\: \\ $$$${solve}\:{for}\:{x}. \\ $$ Commented by Prithwish sen last updated on 30/Jul/19 $$\mathrm{x}^{\mathrm{3}}…
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Question Number 65463 by aliesam last updated on 30/Jul/19 Answered by MJS last updated on 31/Jul/19 $$\mathrm{i}^{{an}} =\mathrm{cos}\:\frac{{an}\pi}{\mathrm{2}}\:+\mathrm{i}\:\mathrm{sin}\:\frac{{an}\pi}{\mathrm{2}} \\ $$$$\mathrm{imag}\:\left(\mathrm{i}^{{n}} +\mathrm{i}^{\mathrm{2}{n}} +\mathrm{i}^{\mathrm{3}{n}} \right)\:=\mathrm{0} \\ $$$$\mathrm{sin}\:\frac{{n}\pi}{\mathrm{2}}\:+\mathrm{sin}\:{n}\pi\:+\mathrm{sin}\:\frac{\mathrm{3}{n}\pi}{\mathrm{2}}\:=\mathrm{0}…
Question Number 65457 by behi83417@gmail.com last updated on 30/Jul/19 $$\begin{cases}{\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}+\frac{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}}=\mathrm{2}\sqrt{\mathrm{3}}}\\{\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}}+\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{a}}}=\mathrm{3}\sqrt{\mathrm{2}}}\end{cases}\:\:\:\:\:\left[\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\boldsymbol{\mathrm{R}},\boldsymbol{\mathrm{a}}\neq\boldsymbol{\mathrm{b}}\right] \\ $$ Answered by mr W last updated on 30/Jul/19 $${let}\:{t}=\frac{{a}}{{b}} \\ $$$${eqn}.\:\mathrm{1}\:\Rightarrow{t}+\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}=\mathrm{2}\sqrt{\mathrm{3}}\:\:\:\:\:\:…\left({i}\right) \\ $$$${eqn}.\:\mathrm{2}\:\Rightarrow\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}=\mathrm{3}\sqrt{\mathrm{2}}\:\:\:\:…\left({ii}\right)…
Question Number 65347 by Masumsiddiqui399@gmail.com last updated on 28/Jul/19 Answered by MJS last updated on 28/Jul/19 $$\mathrm{squaring}\:\left(\Rightarrow\:\mathrm{beware}\:\mathrm{of}\:\mathrm{false}\:\mathrm{solutions}!\right) \\ $$$$\Rightarrow \\ $$$$\left({x}^{\mathrm{2}} +{x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{0} \\…
Question Number 130853 by Koyoooo last updated on 29/Jan/21 Answered by mathmax by abdo last updated on 29/Jan/21 $$\mathrm{for}\:\mathrm{x}\in\mathrm{C}\:\:\:\:\mathrm{no}\:\mathrm{difference}\:\:\sqrt{\mathrm{x}^{\mathrm{2}} }=\left(\sqrt{\mathrm{x}}\right)^{\mathrm{2}} \\ $$$$\mathrm{inside}\:\mathrm{R}\:\:\sqrt{\mathrm{x}^{\mathrm{2}} }=\mid\mathrm{x}\mid\:\mathrm{here}\:\mathrm{x}\:\in\mathrm{R}\:\:\mathrm{and}\:\left(\sqrt{\mathrm{x}}\right)^{\mathrm{2}} \:=\mathrm{x}\:\mathrm{here}\:\mathrm{x}\in\mathrm{R}^{+} \\…
Question Number 130846 by Eric002 last updated on 29/Jan/21 $${prove}\:{that} \\ $$$$\mathrm{3}{sec}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)−\mathrm{4}{csc}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)+\mathrm{5}{cot}^{−\mathrm{1}} \left(\mathrm{2}\right)=\mathrm{1}.\mathrm{533} \\ $$ Commented by MJS_new last updated on 30/Jan/21 $$\mathrm{we}\:\mathrm{cannot}\:\mathrm{prove}\:\mathrm{this}\:\mathrm{as}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{true}…
Question Number 65271 by ajfour last updated on 27/Jul/19 $${x}^{\mathrm{4}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${solve}\:{for}\:{x}. \\ $$ Answered by ajfour last updated on 29/Jul/19 $${let}\:{x}^{\mathrm{2}} ={px}+{t}…
Question Number 130806 by shaker last updated on 29/Jan/21 Answered by Olaf last updated on 29/Jan/21 $$\mathrm{sh}{x}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:=\:\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}} \\ $$$${e}^{\mathrm{2}{x}} −\frac{\mathrm{4}}{\mathrm{3}}{e}^{{x}} −\mathrm{1}\:=\:\mathrm{0} \\ $$$${e}^{{x}}…
Question Number 130800 by ajfour last updated on 29/Jan/21 Commented by ajfour last updated on 29/Jan/21 $${Q}.\mathrm{130358} \\ $$ Terms of Service Privacy Policy Contact:…