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Question Number 12682 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 28/Apr/17

Answered by mrW1 last updated on 29/Apr/17

let u=x+y  let v=xy    (x+y)^2 =x^2 +y^2 +2xy=4+2xy  ⇒u^2 =4+2v   ...(i)    x^3 +y^3 =(x+y)(x^2 +y^2 −xy)=8  (x+y)(4−xy)=8  ⇒u(4−v)=8    ...(ii)    from (i):  v=(u^2 /2)−2  into (ii):  u(4−(u^2 /2)+2)=8  u(12−u^2 )=16  u^3 −12u+16=0  (u−2)(u−2)(u+4)=0  ⇒u=2 or −4  ⇒v=0 or 6    ⇒x+y=2  ⇒xy=0  ⇒x^2 +y^2 −2xy=4  ⇒(x−y)^2 =4  ⇒x−y=±2  ⇒x=2 or 0  ⇒y=0 or 2    ⇒x+y=−4  ⇒xy=6  ⇒x^2 +y^2 −2xy=4−12=−8  ⇒(x−y)^2 =−8  ⇒x−y=±2(√2)i  ⇒x=−2±(√2)i  ⇒y=−2∓(√2)i    so the solutions are:   ((x),(y) )= ((2),(0) ) or  ((0),(2) ) or  (((−2+(√2)i)),((−2−(√2)i)) ) or  (((−2−(√2)i)),((−2+(√2)i)) )

$${let}\:{u}={x}+{y} \\ $$$${let}\:{v}={xy} \\ $$$$ \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{4}+\mathrm{2}{xy} \\ $$$$\Rightarrow{u}^{\mathrm{2}} =\mathrm{4}+\mathrm{2}{v}\:\:\:...\left({i}\right) \\ $$$$ \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}\right)=\mathrm{8} \\ $$$$\left({x}+{y}\right)\left(\mathrm{4}−{xy}\right)=\mathrm{8} \\ $$$$\Rightarrow{u}\left(\mathrm{4}−{v}\right)=\mathrm{8}\:\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right): \\ $$$${v}=\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2} \\ $$$${into}\:\left({ii}\right): \\ $$$${u}\left(\mathrm{4}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}\right)=\mathrm{8} \\ $$$${u}\left(\mathrm{12}−{u}^{\mathrm{2}} \right)=\mathrm{16} \\ $$$${u}^{\mathrm{3}} −\mathrm{12}{u}+\mathrm{16}=\mathrm{0} \\ $$$$\left({u}−\mathrm{2}\right)\left({u}−\mathrm{2}\right)\left({u}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{2}\:{or}\:−\mathrm{4} \\ $$$$\Rightarrow{v}=\mathrm{0}\:{or}\:\mathrm{6} \\ $$$$ \\ $$$$\Rightarrow{x}+{y}=\mathrm{2} \\ $$$$\Rightarrow{xy}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{4} \\ $$$$\Rightarrow\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow{x}−{y}=\pm\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{2}\:{or}\:\mathrm{0} \\ $$$$\Rightarrow{y}=\mathrm{0}\:{or}\:\mathrm{2} \\ $$$$ \\ $$$$\Rightarrow{x}+{y}=−\mathrm{4} \\ $$$$\Rightarrow{xy}=\mathrm{6} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{4}−\mathrm{12}=−\mathrm{8} \\ $$$$\Rightarrow\left({x}−{y}\right)^{\mathrm{2}} =−\mathrm{8} \\ $$$$\Rightarrow{x}−{y}=\pm\mathrm{2}\sqrt{\mathrm{2}}{i} \\ $$$$\Rightarrow{x}=−\mathrm{2}\pm\sqrt{\mathrm{2}}{i} \\ $$$$\Rightarrow{y}=−\mathrm{2}\mp\sqrt{\mathrm{2}}{i} \\ $$$$ \\ $$$${so}\:{the}\:{solutions}\:{are}: \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}}\\{\mathrm{0}}\end{pmatrix}\:{or}\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{2}}\end{pmatrix}\:{or}\:\begin{pmatrix}{−\mathrm{2}+\sqrt{\mathrm{2}}{i}}\\{−\mathrm{2}−\sqrt{\mathrm{2}}{i}}\end{pmatrix}\:{or}\:\begin{pmatrix}{−\mathrm{2}−\sqrt{\mathrm{2}}{i}}\\{−\mathrm{2}+\sqrt{\mathrm{2}}{i}}\end{pmatrix} \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/Apr/17

thanks a lot dear mrW1.you are no.1.

$${thanks}\:{a}\:{lot}\:{dear}\:{mrW}\mathrm{1}.{you}\:{are}\:{no}.\mathrm{1}. \\ $$

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