Question Number 49186 by afachri last updated on 04/Dec/18 $$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{help}}… \\ $$$$ \\ $$$$\mathrm{There}\:\mathrm{is}\::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{21}\boldsymbol{{x}}^{\mathrm{2}} \:−\:\mathrm{21}\boldsymbol{{p}}\:\boldsymbol{{x}}\:+\:\mathrm{49}\boldsymbol{{p}}\:−\:\mathrm{7}\:=\:\mathrm{0} \\ $$$$\mathrm{whose}\:\mathrm{roots}\:\boldsymbol{{u}}\:\mathrm{and}\:\boldsymbol{{v}}.\:\mathrm{If}\:\boldsymbol{{u}}\:\mathrm{and}\:\boldsymbol{{v}}\:\mathrm{are}\:\mathrm{not}\:\in\mathbb{Z}\:,\: \\ $$$$\mathrm{and}\:\boldsymbol{{u}},\boldsymbol{{v}}\:\geqslant\:\mathrm{1}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\boldsymbol{{u}}\:+\:\boldsymbol{{v}}\:! \\ $$$$…
Question Number 114720 by Algoritm last updated on 20/Sep/20 Answered by 1549442205PVT last updated on 20/Sep/20 $$\mathrm{x}^{\mathrm{4}} −\mathrm{2x}^{\mathrm{3}} +\mathrm{4x}−\mathrm{2}\Leftrightarrow\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{6x}−\mathrm{3}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 49177 by rahul 19 last updated on 04/Dec/18 $${If}\:\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{1}} \mid+\mid{z}_{\mathrm{2}} \mid\:,\:{then}\:{prove}\: \\ $$$${that}\:{arg}\left(\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }\right)=\pi\:. \\ $$ Answered by MJS last…
Question Number 49174 by rahul 19 last updated on 04/Dec/18 $${If}\:{z}_{\mathrm{1}} ,{z}_{\mathrm{2}} \:{and}\:{z}_{\mathrm{3}} ,{z}_{\mathrm{4}\:} {are}\:{two}\:{pairs}\:{of}\: \\ $$$${conjugate}\:{complex}\:{numbers}\:,\:{then}\: \\ $$$${find}\:{value}\:{of}\:{arg}\left(\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{4}} }\right)+{arg}\left(\frac{{z}_{\mathrm{2}} }{{z}_{\mathrm{3}} }\right)\:? \\ $$…
Question Number 180236 by mnjuly1970 last updated on 09/Nov/22 $$\:\:\:\:\mathrm{calculation} \\ $$$$\:\:\:\:\Omega=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\zeta\:\left(\:\mathrm{2}{n}\:\right)}{\mathrm{4}^{\:{n}} }\:\overset{?} {=}\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\:\Omega\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{\frac{\mathrm{1}}{\mathrm{2}^{\:\mathrm{2}{n}} }\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\:\mathrm{2}{n}} }\:\right\} \\…
Question Number 49169 by rahul 19 last updated on 04/Dec/18 $${Find}\:{the}\:{least}\:{positive}\:{integer}\:{n}\:{such} \\ $$$${that}\:\left(\frac{\mathrm{2}{i}}{\mathrm{1}+{i}}\right)^{{n}} {is}\:{a}\:+{ve}\:{integer}\:? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18 $$\left(\frac{\mathrm{2}{i}\left(\mathrm{1}−{i}\right)}{\mathrm{2}}\right)^{{n}} \\…
Question Number 49151 by rahul 19 last updated on 03/Dec/18 $${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 Number 49147 by behi83417@gmail.com last updated on 03/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 03/Dec/18 $${ab}−{ad}+{cd}−{bc}=\mathrm{0} \\ $$$${a}\left({b}−{d}\right)−{c}\left({b}−{d}\right)=\mathrm{0} \\ $$$$\left({a}−{c}\right)\left({b}−{d}\right)=\mathrm{0} \\ $$$${either}\:{a}={c}\:\:{or}\:{b}={d} \\ $$$${let}\:{a}={c}…
Question Number 180211 by Shrinava last updated on 09/Nov/22 $$\mathrm{Find}: \\ $$$$\Omega\:=\:\mathrm{gcd}\:\left(\mathrm{4}^{\mathrm{2020}} −\mathrm{1}\:,\:\mathrm{8}^{\mathrm{2021}} −\mathrm{1}\right).\:\mathrm{Generalization}. \\ $$ Answered by puissant last updated on 09/Nov/22 $$\Omega\:=\:{gcd}\left(\mathrm{2}^{\mathrm{4040}} −\mathrm{1}\:;\:\mathrm{2}^{\mathrm{6063}}…
Question Number 180207 by Acem last updated on 09/Nov/22 $$\:\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}\:\: \\…