Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 84770 by jagoll last updated on 16/Mar/20

x^2 +y^2  = 30   (1/x)+(1/y) = 2   find the solution x & y ?

$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{30}\: \\ $$ $$\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}}\:=\:\mathrm{2}\: \\ $$ $$\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{x}\:\&\:\mathrm{y}\:? \\ $$

Commented byTony 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)

$$\frac{{x}+{y}}{{xy}}=\mathrm{2} \\ $$ $${x}+{y}=\mathrm{2}{xy} \\ $$ $${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$ $$=\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy} \\ $$ $$=\left({x}+{y}\right)^{\mathrm{2}} −\left({x}+{y}\right)=\mathrm{30} \\ $$ $$\left[\left({x}+{y}\right)−\mathrm{6}\right]\left[\left({x}+{y}\right)+\mathrm{5}\right]=\mathrm{0} \\ $$ $$\Rightarrow\begin{cases}{{x}+{y}=\mathrm{6}}\\{{xy}=\mathrm{3}}\end{cases}\:\:\:\:\:\:\:\:\begin{cases}{{x}+{y}=−\mathrm{5}}\\{{xy}=−\frac{\mathrm{5}}{\mathrm{2}}}\end{cases} \\ $$ $${x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{3}=\mathrm{0} \\ $$ $${x}=\mathrm{3}\pm\sqrt{\mathrm{6}} \\ $$ $${y}=\frac{\mathrm{3}}{{x}}=\mathrm{3}\mp\sqrt{\mathrm{6}} \\ $$ $$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{5}=\mathrm{0} \\ $$ $${x}=\frac{−\mathrm{5}\pm\sqrt{\mathrm{35}}}{\mathrm{2}} \\ $$ $${y}=\frac{−\mathrm{5}\mp\sqrt{\mathrm{35}}}{\mathrm{2}} \\ $$

Answered by jagoll last updated on 16/Mar/20

Answered by behi83417@gmail.com last updated on 16/Mar/20

x+y=p,xy=q⇒ { ((p^2 −2q=30)),(((p/q)=2⇒p=2q)) :}  ⇒(2q)^2 −2q−30=0⇒2q=((1±(√(121)))/2)  ⇒2q=6,−5⇒ { ((q=3,−(5/2))),((p=6,−5)) :}  ⇒z^2 −6z+3=0⇒z_(1,2) =x,y=((6±(√(24)))/2)=3±(√6)  ⇒z^2 +5z−(5/2)=0⇒z_(3,4) =x,y=((−5±(√(35)))/2)

$$\mathrm{x}+\mathrm{y}=\mathrm{p},\mathrm{xy}=\mathrm{q}\Rightarrow\begin{cases}{\mathrm{p}^{\mathrm{2}} −\mathrm{2q}=\mathrm{30}}\\{\frac{\mathrm{p}}{\mathrm{q}}=\mathrm{2}\Rightarrow\mathrm{p}=\mathrm{2q}}\end{cases} \\ $$ $$\Rightarrow\left(\mathrm{2q}\right)^{\mathrm{2}} −\mathrm{2q}−\mathrm{30}=\mathrm{0}\Rightarrow\mathrm{2q}=\frac{\mathrm{1}\pm\sqrt{\mathrm{121}}}{\mathrm{2}} \\ $$ $$\Rightarrow\mathrm{2q}=\mathrm{6},−\mathrm{5}\Rightarrow\begin{cases}{\mathrm{q}=\mathrm{3},−\frac{\mathrm{5}}{\mathrm{2}}}\\{\mathrm{p}=\mathrm{6},−\mathrm{5}}\end{cases} \\ $$ $$\Rightarrow\mathrm{z}^{\mathrm{2}} −\mathrm{6z}+\mathrm{3}=\mathrm{0}\Rightarrow\mathrm{z}_{\mathrm{1},\mathrm{2}} =\mathrm{x},\mathrm{y}=\frac{\mathrm{6}\pm\sqrt{\mathrm{24}}}{\mathrm{2}}=\mathrm{3}\pm\sqrt{\mathrm{6}} \\ $$ $$\Rightarrow\mathrm{z}^{\mathrm{2}} +\mathrm{5z}−\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{0}\Rightarrow\mathrm{z}_{\mathrm{3},\mathrm{4}} =\mathrm{x},\mathrm{y}=\frac{−\mathrm{5}\pm\sqrt{\mathrm{35}}}{\mathrm{2}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com