Question Number 84770 by jagoll last updated on 16/Mar/20
$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{30}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}}\:=\:\mathrm{2}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{x}\:\&\:\mathrm{y}\:? \\ $$
Commented by Tony Lin last updated on 16/Mar/20
![((x+y)/(xy))=2 x+y=2xy x^2 +y^2 =(x+y)^2 −2xy =(x+y)^2 −(x+y)=30 [(x+y)−6][(x+y)+5]=0 ⇒ { ((x+y=6)),((xy=3)) :} { ((x+y=−5)),((xy=−(5/2))) :} x^2 −6x+3=0 x=3±(√6) y=(3/x)=3∓(√6) 2x^2 +10x−5=0 x=((−5±(√(35)))/2) y=((−5∓(√(35)))/2)](https://www.tinkutara.com/question/Q84773.png)
$$\frac{{x}+{y}}{{xy}}=\mathrm{2} \\ $$$${x}+{y}=\mathrm{2}{xy} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$$=\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy} \\ $$$$=\left({x}+{y}\right)^{\mathrm{2}} −\left({x}+{y}\right)=\mathrm{30} \\ $$$$\left[\left({x}+{y}\right)−\mathrm{6}\right]\left[\left({x}+{y}\right)+\mathrm{5}\right]=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{x}+{y}=\mathrm{6}}\\{{xy}=\mathrm{3}}\end{cases}\:\:\:\:\:\:\:\:\begin{cases}{{x}+{y}=−\mathrm{5}}\\{{xy}=−\frac{\mathrm{5}}{\mathrm{2}}}\end{cases} \\ $$$${x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{3}=\mathrm{0} \\ $$$${x}=\mathrm{3}\pm\sqrt{\mathrm{6}} \\ $$$${y}=\frac{\mathrm{3}}{{x}}=\mathrm{3}\mp\sqrt{\mathrm{6}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{5}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{5}\pm\sqrt{\mathrm{35}}}{\mathrm{2}} \\ $$$${y}=\frac{−\mathrm{5}\mp\sqrt{\mathrm{35}}}{\mathrm{2}} \\ $$
Answered by jagoll last updated on 16/Mar/20
Answered by behi83417@gmail.com last updated on 16/Mar/20
$$\mathrm{x}+\mathrm{y}=\mathrm{p},\mathrm{xy}=\mathrm{q}\Rightarrow\begin{cases}{\mathrm{p}^{\mathrm{2}} −\mathrm{2q}=\mathrm{30}}\\{\frac{\mathrm{p}}{\mathrm{q}}=\mathrm{2}\Rightarrow\mathrm{p}=\mathrm{2q}}\end{cases} \\ $$$$\Rightarrow\left(\mathrm{2q}\right)^{\mathrm{2}} −\mathrm{2q}−\mathrm{30}=\mathrm{0}\Rightarrow\mathrm{2q}=\frac{\mathrm{1}\pm\sqrt{\mathrm{121}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2q}=\mathrm{6},−\mathrm{5}\Rightarrow\begin{cases}{\mathrm{q}=\mathrm{3},−\frac{\mathrm{5}}{\mathrm{2}}}\\{\mathrm{p}=\mathrm{6},−\mathrm{5}}\end{cases} \\ $$$$\Rightarrow\mathrm{z}^{\mathrm{2}} −\mathrm{6z}+\mathrm{3}=\mathrm{0}\Rightarrow\mathrm{z}_{\mathrm{1},\mathrm{2}} =\mathrm{x},\mathrm{y}=\frac{\mathrm{6}\pm\sqrt{\mathrm{24}}}{\mathrm{2}}=\mathrm{3}\pm\sqrt{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{z}^{\mathrm{2}} +\mathrm{5z}−\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{0}\Rightarrow\mathrm{z}_{\mathrm{3},\mathrm{4}} =\mathrm{x},\mathrm{y}=\frac{−\mathrm{5}\pm\sqrt{\mathrm{35}}}{\mathrm{2}} \\ $$