Category: Algebra

Question Number 210996 by RojaTaniya last updated on 25/Aug/24 Answered by Frix last updated on 26/Aug/24 $$\mathrm{Only}\:\mathrm{true}\:\mathrm{for}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solution}\:>\mathrm{0}: \\ $$$${x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{9}{x}=\mathrm{4}{x}^{\mathrm{2}}…

Question Number 210961 by RojaTaniya last updated on 24/Aug/24 Answered by A5T last updated on 24/Aug/24 $$\frac{{a}−\mathrm{1}+\mathrm{2}}{{a}−\mathrm{1}}+\frac{{b}−\mathrm{1}+\mathrm{2}}{{b}−\mathrm{1}}+\frac{{c}−\mathrm{1}+\mathrm{2}}{{c}−\mathrm{1}}=\mathrm{10} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}−\mathrm{1}}+\frac{\mathrm{1}}{{b}−\mathrm{1}}+\frac{\mathrm{1}}{{c}−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${a}+{b}+{c}=\mathrm{0};{ab}+{bc}+{ca}=\frac{{m}−\mathrm{1}}{{m}};{abc}=\frac{−{m}−\mathrm{1}}{{m}} \\ $$$$\Rightarrow\frac{{ab}+{bc}+{ca}−\mathrm{2}\left({a}+{b}+{c}\right)+\mathrm{3}}{{abc}−{ab}−{bc}−{ac}+{a}+{b}+{c}−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\frac{{m}−\mathrm{1}+\mathrm{3}{m}}{{m}}}{\frac{−{m}−\mathrm{1}+\mathrm{1}−{m}−{m}}{{m}}}=\frac{\mathrm{7}}{\mathrm{2}}\Rightarrow\frac{\mathrm{4}{m}−\mathrm{1}}{−\mathrm{3}{m}}=\frac{\mathrm{7}}{\mathrm{2}}…

Question Number 210934 by hardmath last updated on 22/Aug/24 $$\mathrm{If}\:\:\:\frac{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{ax}\:−\:\mathrm{18}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{7x}\:+\:\mathrm{2b}}\:\:=\:\:\frac{\mathrm{x}\:−\:\mathrm{c}}{\mathrm{x}\:+\:\mathrm{5}} \\ $$$$\mathrm{Find}\:\:\:\boldsymbol{\mathrm{a}}\:+\:\boldsymbol{\mathrm{b}}\:+\:\boldsymbol{\mathrm{c}}\:=\:? \\ $$ Answered by A5T last updated on 22/Aug/24 $$\left({x}+\mathrm{5}\right)\left({x}\right)+\mathrm{2}\left({x}+\mathrm{5}\right)+\mathrm{2}{b}−\mathrm{10}\Rightarrow\mathrm{2}{b}−\mathrm{10}=\mathrm{0}\Rightarrow{b}=\mathrm{5} \\…

Question Number 210906 by hardmath last updated on 21/Aug/24 $$\frac{\mathrm{19x}\:−\:\mathrm{x}^{\mathrm{2}} }{\mathrm{x}\:+\:\mathrm{1}}\:\centerdot\:\left(\mathrm{x}\:+\:\frac{\mathrm{19}\:−\:\mathrm{x}}{\mathrm{x}\:+\:\mathrm{1}}\right)\:=\:\mathrm{78} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by Frix last updated on 21/Aug/24 $$\mathrm{Reconstructing}\:\mathrm{your}\:\mathrm{equation}: \\ $$$${x}=\frac{{a}}{\mathrm{2}}\pm\frac{\sqrt{{a}^{\mathrm{2}}…

Question Number 210873 by hardmath last updated on 20/Aug/24 Terms of Service Privacy Policy Contact: info@tinkutara.com