Question Number 223995 by Rojarani last updated on 13/Aug/25 Answered by Frix last updated on 13/Aug/25 $$\mathrm{Let}\:{y}={px}\wedge{z}={qx},\:\mathrm{solve}\:\mathrm{for}\:{x}^{−\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${x}^{−\mathrm{2}} =\begin{cases}{\mathrm{7}{p}^{\mathrm{2}} +\mathrm{9}{pq}+\mathrm{3}{q}^{\mathrm{2}} +\mathrm{13}{p}+\mathrm{9}{q}+\mathrm{7}}\\{\left(\mathrm{7}{p}^{\mathrm{2}} +\mathrm{13}{pq}+\mathrm{7}{q}^{\mathrm{2}}…
Question Number 223964 by Rojarani last updated on 11/Aug/25 Answered by Rasheed.Sindhi last updated on 11/Aug/25 $${x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{4}=\mathrm{0} \\ $$$${a},{b},{c}\:{are}\:{the}\:{roots} \\ $$$${a}+{b}+{c}=−\mathrm{2} \\ $$$${ab}+{bc}+{ca}=\mathrm{3}…
Question Number 223965 by Rojarani last updated on 11/Aug/25 Answered by Frix last updated on 11/Aug/25 $${z}={x}+{y}−\mathrm{7} \\ $$$$\mathrm{Insert}\:\mathrm{in}\:\left(\mathrm{2}\right)\:\Rightarrow\:{y}=\frac{\mathrm{7}{x}−\mathrm{43}}{{x}−\mathrm{7}} \\ $$$$\mathrm{Insert}\:\mathrm{in}\:\left(\mathrm{3}\right)\:\Rightarrow\:{x}^{\mathrm{2}} −\mathrm{19}{x}+\mathrm{90}=\mathrm{0} \\ $$$$\Rightarrow \\…
Question Number 223858 by Rojarani last updated on 07/Aug/25 Answered by Rasheed.Sindhi last updated on 08/Aug/25 $${let}\:{x}+\mathrm{100}={y} \\ $$$$\frac{\left({y}−\mathrm{2}\right)^{\mathrm{5}} +\left({y}+\mathrm{2}\right)^{\mathrm{5}} }{\left({y}−\mathrm{1}\right)^{\mathrm{5}} +\left({y}+\mathrm{1}\right)^{\mathrm{5}} }=\frac{\mathrm{16}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}}…
Question Number 223822 by fantastic last updated on 06/Aug/25 $$\left(\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} \right) \\ $$$$\:{Rewrite}\:{in}\:{simplest}\:{radical}\:{form} \\ $$ Answered by Raphael254 last updated on 06/Aug/25 $$ \\ $$$$\left(\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\frac{\mathrm{4}}{\mathrm{3}}}…
Question Number 223823 by fantastic last updated on 06/Aug/25 $$\sqrt{\mathrm{4}{x}+\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{2}}=\mathrm{1} \\ $$$${x}=? \\ $$ Answered by Rasheed.Sindhi last updated on 06/Aug/25 $$\sqrt{\mathrm{4}{x}+\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{2}}=\mathrm{1}…{i} \\ $$$$\left(\mathrm{4}{x}+\mathrm{1}\right)−\left(\mathrm{3}{x}−\mathrm{2}\right)=\sqrt{\mathrm{4}{x}+\mathrm{1}}\:−\sqrt{\mathrm{3}{x}−\mathrm{2}}\: \\…
Question Number 223804 by DignoR1 last updated on 05/Aug/25 Commented by Ghisom last updated on 05/Aug/25 $$\mathrm{you}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$ Commented by fantastic last updated on…
Question Number 223800 by efronzo1 last updated on 05/Aug/25 $$\:\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{14x}+\mathrm{40}}{\mathrm{g}\left(\mathrm{x}\right)}−\mathrm{43} \\ $$$$\:\mathrm{h}\left(\mathrm{x}\right)=\:\frac{\mathrm{g}\left(\mathrm{x}\right)+\mathrm{51}}{\mathrm{x}+\mathrm{4}} \\ $$$$\:\mathrm{m}\left(\mathrm{x}\right)=\:\frac{\mathrm{h}\left(\mathrm{x}\right)−\mathrm{9}}{\mathrm{x}−\mathrm{2}}\:,\:\mathrm{x}\neq\mathrm{2} \\ $$$$\:\mathrm{m}\left(\mathrm{2}\right)=\:\mathrm{2043}.\: \\ $$$$\:\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{divided}\:\mathrm{by}\:\mathrm{x}^{\mathrm{2}} +\mathrm{8x}−\mathrm{20}\: \\ $$$$\:\mathrm{gives}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{M}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b} \\ $$$$\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{M}\left(\mathrm{98}\right)=?\: \\…
Question Number 223734 by behi834171 last updated on 03/Aug/25 Answered by Ghisom last updated on 03/Aug/25 $$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$$\mathrm{all}\:\mathrm{paths}\:\mathrm{lead}\:\mathrm{to}\:\mathrm{polynomials}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{8} \\ $$$$\mathrm{I}\:\mathrm{also}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{substitute} \\ $$$${x}=\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nice}\:\mathrm{looking}…
Question Number 223700 by BaliramKumar last updated on 02/Aug/25 $$\:\underline{\underbrace{\:}} \\ $$ Answered by mehdee7396 last updated on 02/Aug/25 $${let}\:\:{a}=\mathrm{2}{k}+\mathrm{1}\Rightarrow{x}^{\mathrm{2}} −\mathrm{2}{kx}+\mathrm{8}{k}+\mathrm{11}=\mathrm{0} \\ $$$$\Rightarrow\Delta={k}^{\mathrm{2}} −\mathrm{8}{k}−\mathrm{11}={n}^{\mathrm{2}} \\…