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Question Number 196968 by sonukgindia last updated on 05/Sep/23

	Answered by MM42 last updated on 05/Sep/23
	$if x,y∈ R { ((5s−p=10+7(√3))),((s^2 −2p=7)) :} ⇒s^2 +10s−27−14(√3)=0 ⇒s=2+(√(3 ))⇒p=2(√3) ⇒X^2 −(2+(√3))X+2(√3)=0 ⇒x=(√(3 ))& y=2$
	$i f x, y \in R$ ${\begin{cases} 5 s - p = 10 + 7 \sqrt{3} \\ s^{2} - 2 p = 7 \end{cases}$ $\Rightarrow s^{2} + 10 s - 27 - 14 \sqrt{3} = 0$ $\Rightarrow s = 2 + \sqrt{3} \Rightarrow p = 2 \sqrt{3}$ $\Rightarrow X^{2} - (2 + \sqrt{3}) X + 2 \sqrt{3} = 0$ $\Rightarrow x = \sqrt{3} & y = 2$
	Answered by Frix last updated on 05/Sep/23

	$a x + x y + a y = b$ $x^{2} + y^{2} = c$ $x = u - v \land y = u + v$ $u^{2} + 2 a u - v^{2} = b \Rightarrow v^{2} = u^{2} + 2 a u - b$ $2 u^{2} + 2 v^{2} = c \Rightarrow v^{2} = \frac{c}{2} - u^{2}$ $u^{2} + 2 a u - b = \frac{c}{2} - u^{2}$ $u^{2} + a u - \frac{2 b + c}{4} = 0$ $u = - \frac{a}{2} \pm \frac{\sqrt{a^{2} + 2 b + c}}{2}$ $v = \sqrt{- \frac{2 a^{2} + 2 b - c}{4} \pm \frac{a \sqrt{a^{2} + 2 b + c}}{2}}$ $With a = 5 \land b = 10 + 7 \sqrt{3} \land c = 7 we get$ $x = \sqrt{3} \land y = 2$ $x = - 6 - \frac{\sqrt{3}}{2} - \frac{\sqrt{33 + 24 \sqrt{3}}}{2} i \land y = - 6 - \frac{\sqrt{3}}{2} + \frac{\sqrt{33 + 24 \sqrt{3}}}{2} i$ $. . . and of course we can exchange x \leftrightarrow y$
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