Question

Salut!
Mă puteți ajuta sa rezolv sistemul acesta de ecuații?
{×+y+5=0
X²-2x+y=3

Answer 1

$\left\{ \begin{array}{ll} x+y+5 = 0\Big|\cdot (-1) \\ x^2-2x+y=3 \end{array} \right \Rightarrow \begin{array}{ll} \left\{ \begin{array}{ll} -x-y-5 = 0 \\ x^2-2x+y=3 \end{array} \\ ----------(+) \end{array}\\ x^2-x-2x-y+y-5 = 3 - 0 \\ x^2-3x -5 - 3 = 0 \\ x^2-3x-8 = 0 \\ \Delta = (-3)^2 - 4\cdot 1\cdot (-8) = 9 + 32 = 41$

$\bullet\left| \begin{array}{ll} x = \dfrac{-(-3) + \sqrt{41}}{2\cdot 1 } = \dfrac{3+\sqrt{41}}{2} \\ \\ \dfrac{3+\sqrt{41}}{2} + y + 5 = 0 \Rightarrow y = - \Big(\dfrac{3+\sqrt{41}}{2} +5 \Big) = -\dfrac{13+\sqrt{41}}{2}\end{array} \right \\ \\ \\\bullet \left|\begin{array}{ll}x = \dfrac{-(-3)-\sqrt{41}}{2\cdot 1} = \dfrac{3-\sqrt{41}}{2} \\ \\\dfrac{3-\sqrt{41}}{2} + y + 5 = 0 \Rightarrow y = -\Big(\dfrac{3-\sqrt{41}}{2} +5 \Big) = \dfrac{\sqrt{41} - 13}{2}\end{array} \right$

$\\ \\ \Rightarrow \boxed{S = \left\{ \left(\dfrac{3+\sqrt{41}}{2}, -\dfrac{13+\sqrt{41}}{2} \right) ; \left( \dfrac{3-\sqrt{41}}{2}, \dfrac{\sqrt{41} - 13}{2}\right) \right\}}$