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Question Number 200265 by cherokeesay last updated on 16/Nov/23

	Answered by witcher3 last updated on 16/Nov/23
	${ (((x^2 −y)^2 +5^2 −2.5(x^2 −y)=0)),(((2)⇔(2))) :} ⇔ { (((x^2 −y−5)^2 =0)),(((√((x^2 −3)^2 +4))−(√(2x^2 +y^2 −6))=0)) :} ⇔ { ((y=x^2 −5)),(((x^4 −6x^2 +13)=2x^2 +y^2 −6)) :} x^4 −8x^2 +19−(x^4 −10x^2 +25)=0 2x^2 −6=0 x=(√3) x=−(√3) y=−2 S={((√3),−2);(−(√3),−2)}$
	${\begin{cases} {(x^{2} - y)}^{2} + 5^{2} - 2.5 (x^{2} - y) = 0 \\ (2) \Leftrightarrow (2) \end{cases}$ $\Leftrightarrow {\begin{cases} {(x^{2} - y - 5)}^{2} = 0 \\ \sqrt{{(x^{2} - 3)}^{2} + 4} - \sqrt{{2 x}^{2} + y^{2} - 6} = 0 \end{cases}$ $\Leftrightarrow {\begin{cases} y = x^{2} - 5 \\ (x^{4} - {6 x}^{2} + 13) = {2 x}^{2} + y^{2} - 6 \end{cases}$ $x^{4} - {8 x}^{2} + 19 - (x^{4} - {10 x}^{2} + 25) = 0$ ${2 x}^{2} - 6 = 0$ $x = \sqrt{3}$ $x = - \sqrt{3}$ $y = - 2$ $S = {(\sqrt{3}, - 2); (- \sqrt{3}, - 2)}$
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