Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
[tex] \left \{ {{y=x-2} \atop {\ y^2=x}} \right. \\ \\ \\
y=x-2 \\
y^2=(x-2)^2=x^2-4x+4 \\ \\ \\
y^2=x \\
x^2-4x+4=x \\
x^2-5x+4=0[/tex]
[tex]a=1 \\ b=-5 \\ c=4 \\ \\ x=\frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}=\frac{5 \pm \sqrt{25 - 16}}{2}=\frac{5 \pm \sqrt{9}}{2}=\frac{5 \pm 3}{2} \\ \\ x=\frac{5+3}{2} \ \lor \ x=\frac{5-3}{2} \\ \\ x=\frac{8}{2} \ \lor \ x=\frac{2}{2} \\ \\ x=4 \ \lor \ x=1[/tex]
[tex]y=4-2 \ \lor \ y=1-2 \\ y=2 \ \lor \ y=-1 \\ \\ \\ \left \{ {{x=4} \atop {y=2}} \right. \ \lor \ \left \{ {{x=1} \atop {y=-1}} \right. [/tex]
[tex]a=1 \\ b=-5 \\ c=4 \\ \\ x=\frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}=\frac{5 \pm \sqrt{25 - 16}}{2}=\frac{5 \pm \sqrt{9}}{2}=\frac{5 \pm 3}{2} \\ \\ x=\frac{5+3}{2} \ \lor \ x=\frac{5-3}{2} \\ \\ x=\frac{8}{2} \ \lor \ x=\frac{2}{2} \\ \\ x=4 \ \lor \ x=1[/tex]
[tex]y=4-2 \ \lor \ y=1-2 \\ y=2 \ \lor \ y=-1 \\ \\ \\ \left \{ {{x=4} \atop {y=2}} \right. \ \lor \ \left \{ {{x=1} \atop {y=-1}} \right. [/tex]
[tex]y=x-2\\
y^2=x\\\\
y=y^2-2\\y^2-y-2=0\\
y^2+y-2y-2=0\\
y(y+1)-2(y+1)=0\\
(y-2)(y+1)=0\\
y=2 \vee y=-1\\\\
x=2^2 \vee x=(-1)^2\\
x=4 \vee x=1\\\\
\boxed{x=4,y=2 \vee x=1,y=-1}
[/tex]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.