Question Number 63373 by MJS last updated on 31/Jul/19 $$\mathrm{just}\:\mathrm{found}\:\mathrm{this}\:\mathrm{on}\:\mathrm{the}\:\mathrm{web} \\ $$$$\mathrm{I}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{might}\:\mathrm{help}\:\mathrm{in}\:\mathrm{some}\:\mathrm{cases}\:\mathrm{where} \\ $$$$\mathrm{quartics}\:\mathrm{appear}\:\mathrm{i}.\mathrm{e}.\:\mathrm{Sir}\:\mathrm{Aifour}'\mathrm{s}\:\mathrm{geometric} \\ $$$$\mathrm{questions}.\:\mathrm{sometimes}\:\mathrm{we}\:\mathrm{know}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{roots},\:\mathrm{but}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{this}\:\mathrm{information}? \\ $$$$ \\ $$$${ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{cx}^{\mathrm{2}} +{dx}+{e}=\mathrm{0}…
Question Number 128903 by Engr_Jidda last updated on 11/Jan/21 $${solve}\:{the}\:{differential}\:{equation} \\ $$$$\frac{{dy}^{\mathrm{2}} }{{dx}^{\mathrm{2}\:} }+{y}=\mathrm{0} \\ $$ Answered by mindispower last updated on 11/Jan/21 $${X}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\Rightarrow{X}\in\left\{{i},−{i}\right\}…
Question Number 128850 by ruwedkabeh last updated on 10/Jan/21 $$\sqrt{{x}+\sqrt{\mathrm{4}{x}+\sqrt{\mathrm{16}{x}+\sqrt{\mathrm{64}{x}+…+\sqrt{\mathrm{4}^{\mathrm{2019}} {x}+\mathrm{3}}}}}}−\sqrt{{x}}=\mathrm{1} \\ $$$${x}=? \\ $$ Answered by mindispower last updated on 10/Jan/21 $$\Rightarrow\sqrt{\mathrm{4}{x}+……+\sqrt{\mathrm{4}^{\mathrm{2009}} {x}+\mathrm{3}}}=\mathrm{1}+\mathrm{2}\sqrt{{x}} \\…
Question Number 63291 by aliesam last updated on 02/Jul/19 $${find}\:{some}\:{of}\:{all}\:{real}\:{x}\:{such}\:{that} \\ $$$$ \\ $$$$\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{17}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{12}}\:=\:\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{16}{x}+\mathrm{18}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{13}} \\ $$ Commented by Prithwish sen last…
Question Number 63289 by Tawa1 last updated on 02/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 63247 by Joel122 last updated on 01/Jul/19 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\sqrt{{abc}}\:+\:\sqrt{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)}\:\leqslant\:\mathrm{1} \\ $$$$\mathrm{for}\:\mathrm{0}\:\leqslant\:{a},{b},{c}\:\leqslant\:\mathrm{1} \\ $$ Commented by Tony Lin last updated on 01/Jul/19 $$\because\:{abc}+\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)\geqslant…
Question Number 128715 by Study last updated on 09/Jan/21 $${fine}\:{the}\:{number}\:{thats}\:{divided}\:{by}\: \\ $$$$\mathrm{3},\mathrm{4},\mathrm{5}\:{and}\:\mathrm{9}\:{and}\:{produce}\:{the}\:{remanider} \\ $$$${with}\:{order}\:\mathrm{1},\mathrm{3},\mathrm{5}\:{and}\:\mathrm{7}?? \\ $$ Commented by mr W last updated on 10/Jan/21 $${do}\:{you}\:{mean}:…
Question Number 128712 by Study last updated on 09/Jan/21 $${if}\:{P}\left({x}−\mathrm{1}\right)=\mathrm{2}{x}^{\mathrm{3}} \centerdot{Q}\left({x}\right)+{x}^{\mathrm{2}} \\ $$$${find}\:\frac{{P}\left(\mathrm{1}\right)−\mathrm{4}}{{Q}\left({x}\right)}=?? \\ $$ Commented by Adel last updated on 13/Jan/21 $$\mathrm{p}\left(\mathrm{1}\right)=\boldsymbol{\mathrm{p}}\left(\mathrm{2}−\mathrm{1}\right)=\mathrm{2}×\mathrm{2}^{\mathrm{3}} ×\boldsymbol{\mathrm{Q}}\left(\mathrm{x}\right)+\mathrm{2}^{\mathrm{2}} \\…
Question Number 63175 by Tawa1 last updated on 30/Jun/19 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \:=\:\:\mathrm{16} \\ $$$$\mathrm{x}\:=\:\mathrm{2},\:\:\:\:\:\mathrm{but}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{Lambert}\:\mathrm{W}\:\mathrm{function} \\ $$ Commented by mr W last updated on 30/Jun/19 $${i}\:{remember}\:{you}\:{have}\:{asked}\:{the}\:{same}…
Question Number 63162 by Rio Michael last updated on 29/Jun/19 $${Find}\:{the}\:{set}\:{of}\:{values}\:{of}\:{x}\:{which}\:{satisfy}\:{the}\:{inequalities}\: \\ $$$$\frac{\mathrm{2}}{{x}−\mathrm{1}}\leqslant\frac{\mathrm{1}}{{x}}\:\:{and}\:\:{x}^{\mathrm{2}} −\mid\mathrm{3}{x}\mid+\mathrm{2}<\mathrm{0} \\ $$ Commented by Prithwish sen last updated on 30/Jun/19 $$\frac{\mathrm{2}}{\mathrm{x}−\mathrm{1}}\leqslant\frac{\mathrm{1}}{\mathrm{x}}\:\Rightarrow\mathrm{x}\leqslant−\mathrm{1}…