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Question-196968

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Question Number 196968 by sonukgindia last updated on 05/Sep/23
Answered by MM42 last updated on 05/Sep/23
if x,y∈ R { ((5s−p=10+7(√3))),((s^2 −2p=7)) :} ⇒s^2 +10s−27−14(√3)=0 ⇒s=2+(√(3 ))⇒p=2(√3) ⇒X^2 −(2+(√3))X+2(√3)=0 ⇒x=(√(3 ))& y=2
$${if}\:\:{x},{y}\in\:\mathbb{R} \\ $$$$\begin{cases}{\mathrm{5}{s}−{p}=\mathrm{10}+\mathrm{7}\sqrt{\mathrm{3}}}\\{{s}^{\mathrm{2}} −\mathrm{2}{p}=\mathrm{7}}\end{cases} \\ $$$$\Rightarrow{s}^{\mathrm{2}} +\mathrm{10}{s}−\mathrm{27}−\mathrm{14}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow{s}=\mathrm{2}+\sqrt{\mathrm{3}\:}\Rightarrow{p}=\mathrm{2}\sqrt{\mathrm{3}}\: \\ $$$$\Rightarrow{X}^{\mathrm{2}} −\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){X}+\mathrm{2}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{3}\:\:}\&\:{y}=\mathrm{2} \\ $$$$ \\ $$
Answered by Frix last updated on 05/Sep/23
ax+xy+ay=b x^2 +y^2 =c x=u−v∧y=u+v u^2 +2au−v^2 =b ⇒ v^2 =u^2 +2au−b 2u^2 +2v^2 =c ⇒ v^2 =(c/2)−u^2 u^2 +2au−b=(c/2)−u^2 u^2 +au−((2b+c)/4)=0 u=−(a/2)±((√(a^2 +2b+c))/2) v=(√(−((2a^2 +2b−c)/4)±((a(√(a^2 +2b+c)))/2))) With a=5∧b=10+7(√3)∧c=7 we get x=(√3)∧y=2 x=−6−((√3)/2)−((√(33+24(√3)))/2)i∧y=−6−((√3)/2)+((√(33+24(√3)))/2)i ...and of course we can exchange x↔y
$${ax}+{xy}+{ay}={b} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={c} \\ $$$${x}={u}−{v}\wedge{y}={u}+{v} \\ $$$${u}^{\mathrm{2}} +\mathrm{2}{au}−{v}^{\mathrm{2}} ={b}\:\Rightarrow\:{v}^{\mathrm{2}} ={u}^{\mathrm{2}} +\mathrm{2}{au}−{b} \\ $$$$\mathrm{2}{u}^{\mathrm{2}} +\mathrm{2}{v}^{\mathrm{2}} ={c}\:\Rightarrow\:{v}^{\mathrm{2}} =\frac{{c}}{\mathrm{2}}−{u}^{\mathrm{2}} \\ $$$${u}^{\mathrm{2}} +\mathrm{2}{au}−{b}=\frac{{c}}{\mathrm{2}}−{u}^{\mathrm{2}} \\ $$$${u}^{\mathrm{2}} +{au}−\frac{\mathrm{2}{b}+{c}}{\mathrm{4}}=\mathrm{0} \\ $$$${u}=−\frac{{a}}{\mathrm{2}}\pm\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{b}+{c}}}{\mathrm{2}} \\ $$$${v}=\sqrt{−\frac{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}−{c}}{\mathrm{4}}\pm\frac{{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{b}+{c}}}{\mathrm{2}}} \\ $$$$ \\ $$$$\mathrm{With}\:{a}=\mathrm{5}\wedge{b}=\mathrm{10}+\mathrm{7}\sqrt{\mathrm{3}}\wedge{c}=\mathrm{7}\:\mathrm{we}\:\mathrm{get} \\ $$$${x}=\sqrt{\mathrm{3}}\wedge{y}=\mathrm{2} \\ $$$${x}=−\mathrm{6}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{33}+\mathrm{24}\sqrt{\mathrm{3}}}}{\mathrm{2}}\mathrm{i}\wedge{y}=−\mathrm{6}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{33}+\mathrm{24}\sqrt{\mathrm{3}}}}{\mathrm{2}}\mathrm{i} \\ $$$$…\mathrm{and}\:\mathrm{of}\:\mathrm{course}\:\mathrm{we}\:\mathrm{can}\:\mathrm{exchange}\:{x}\leftrightarrow{y} \\ $$

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