Find all pair(x,y) of real numbers that are the solutions to the system { ((x^4 +2x^3 −y=−(1/4)+(√3))),((y^4 +2y^3 −x=−(1/4)−(√3))) :}


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Question Number 119849 by benjo_mathlover last updated on 27/Oct/20
$Find all pair(x,y) of real numbers that are the solutions to the system { ((x^4 +2x^3 −y=−(1/4)+(√3))),((y^4 +2y^3 −x=−(1/4)−(√3))) :}$
$F i n d a l l p a i r (x, y) o f r e a l n u m b e r s$ $t h a t a r e t h e s o l u t i o n s t o t h e s y s t e m$ ${\begin{cases} x^{4} + 2 x^{3} - y = - \frac{1}{4} + \sqrt{3} \\ y^{4} + 2 y^{3} - x = - \frac{1}{4} - \sqrt{3} \end{cases}$
	Answered by 1549442205PVT last updated on 27/Oct/20
	$Adding up two the equations we get x^4 +2x^3 −y+(1/4)+y^4 +2y^3 −x+(1/4)=0 ⇔(x^2 +x−(1/2))^2 +(y^2 +y−(1/2))^2 =0 ⇔ { ((x^2 +x−(1/2)=0)),((y^2 +y−(1/2)=0)) :} Δ=1+4.0.5=3,so x= { ((x=((−1±(√3))/2))),((y=((−1±(√3))/2))) :} Thus the solutions of given system are (x,y)∈{(((−1−(√3))/2),((−1−(√3))/2)),(((−1−(√3))/2),((−1+(√3))/2)) ,(((−1+(√3))/2),((−1−(√3))/2)),(((−1+(√3))/2),((−1+(√3))/2))}$
	$Adding up two the equations we get$ $x^{4} + 2 x^{3} - y + \frac{1}{4} + y^{4} + 2 y^{3} - x + \frac{1}{4} = 0$ $\Leftrightarrow {(x^{2} + x - \frac{1}{2})}^{2} + {(y^{2} + y - \frac{1}{2})}^{2} = 0$ $\Leftrightarrow {\begin{cases} x^{2} + x - \frac{1}{2} = 0 \\ y^{2} + y - \frac{1}{2} = 0 \end{cases}$ $Δ = 1 + 4.0.5 = 3, so x = {\begin{cases} x = \frac{- 1 \pm \sqrt{3}}{2} \\ y = \frac{- 1 \pm \sqrt{3}}{2} \end{cases}$ $Thus the solutions of given system are$ $(x, y) \in {(\frac{- 1 - \sqrt{3}}{2}, \frac{- 1 - \sqrt{3}}{2}), (\frac{- 1 - \sqrt{3}}{2}, \frac{- 1 + \sqrt{3}}{2})$ $, (\frac{- 1 + \sqrt{3}}{2}, \frac{- 1 - \sqrt{3}}{2}), (\frac{- 1 + \sqrt{3}}{2}, \frac{- 1 + \sqrt{3}}{2})}$
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