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Category: Algebra

Let-z-is-complex-number-satisfying-the-equation-z-2-3-i-z-m-2i-0-where-m-R-Suppose-the-equation-has-a-real-root-then-find-the-non-real-root-

Question Number 49151 by rahul 19 last updated on 03/Dec/18 $${Let}\:{z}\:{is}\:{complex}\:{number}\:{satisfying} \\ $$$${the}\:{equation}\:{z}^{\mathrm{2}} −\left(\mathrm{3}+{i}\right){z}+{m}+\mathrm{2}{i}=\mathrm{0}, \\ $$$${where}\:{m}\epsilon{R}.\:{Suppose}\:{the}\:{equation} \\ $$$${has}\:{a}\:{real}\:{root},\:{then}\:{find}\:{the}\:{non}\:{real}\:{root}? \\ $$ Answered by mr W last…

Question-49147

Question Number 49147 by behi83417@gmail.com last updated on 03/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 03/Dec/18 $${ab}−{ad}+{cd}−{bc}=\mathrm{0} \\ $$$${a}\left({b}−{d}\right)−{c}\left({b}−{d}\right)=\mathrm{0} \\ $$$$\left({a}−{c}\right)\left({b}−{d}\right)=\mathrm{0} \\ $$$${either}\:{a}={c}\:\:{or}\:{b}={d} \\ $$$${let}\:{a}={c}…

2-15-11-40-26-75-47-120-

Question Number 180207 by Acem last updated on 09/Nov/22 $$\:\frac{\mathrm{2}}{\mathrm{15}}\:,\:\:\frac{\mathrm{11}}{\mathrm{40}}\:\:,\:\:\frac{\mathrm{26}}{\mathrm{75}}\:\:,\:\:\frac{\mathrm{47}}{\mathrm{120}}\:\:,\:… \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last updated on 09/Nov/22 $${Numerators}: \\ $$$$\begin{bmatrix}{\mathrm{2}}&{\:}&{\mathrm{11}}&{\:}&{\mathrm{26}}&{\:}&{\mathrm{47}}&{\:}&{\mathrm{74}}\\{\:}&{\mathrm{9}}&{\:}&{\mathrm{15}}&{\:}&{\mathrm{21}}&{\:}&{\mathrm{27}}&{\:}\\{\:}&{\:}&{\:\mathrm{6}}&{\:}&{\mathrm{6}}&{\:}&{\mathrm{6}}&{\:}&{\:}\end{bmatrix}\:\: \\…

Question-180188

Question Number 180188 by Shrinava last updated on 08/Nov/22 Answered by Rasheed.Sindhi last updated on 09/Nov/22 $$\mathcal{F}{ormulas}: \\ $$$$\begin{array}{|c|}{\:\:\begin{bmatrix}{\:\:\:\:\:{l}}&{{m}}\\{−{m}}&{{l}}\end{bmatrix}\begin{bmatrix}{\:\:\:\:{p}}&{{q}}\\{−{q}}&{{p}}\end{bmatrix}=\begin{bmatrix}{\:\:\:\:\:\:{lp}−{mq}}&{{lq}+{mp}}\\{−\left({lq}+{mp}\right)}&{{lp}−{mq}}\end{bmatrix}_{\:} ^{\:} }\\\hline\end{array}\: \\ $$$$\begin{array}{|c|}{\left(\begin{bmatrix}{\:\:\:\:\:{l}}&{{m}}\\{−{m}}&{{l}}\end{bmatrix}\right)^{\mathrm{2}} \:=\begin{bmatrix}{{l}^{\mathrm{2}} −{m}^{\mathrm{2}}…

solve-6x-4-25x-3-12x-2-25x-6-0-

Question Number 114653 by bemath last updated on 20/Sep/20 $${solve}\:\mathrm{6}{x}^{\mathrm{4}} −\mathrm{25}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} −\mathrm{25}{x}+\mathrm{6}=\mathrm{0} \\ $$ Answered by bobhans last updated on 20/Sep/20 $$\Leftrightarrow\mathrm{6}\left({x}^{\mathrm{4}} +\mathrm{1}\right)−\mathrm{25}\left({x}^{\mathrm{3}} +{x}\right)+\mathrm{12}{x}^{\mathrm{2}}…

Question-114625

Question Number 114625 by bobhans last updated on 20/Sep/20 Answered by john santu last updated on 20/Sep/20 $$\Leftrightarrow{k}^{\mathrm{4}} +\mathrm{2}{k}^{\mathrm{2}} +\mathrm{9}={k}^{\mathrm{4}} +\mathrm{6}{k}^{\mathrm{2}} +\mathrm{9}−\mathrm{4}{k}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{3}\right)\left({k}^{\mathrm{2}}…