Question

Solve simultaneously Y=x-2 Y^2=x

Answer 1

$\left \{ {{y=x-2} \atop {\ y^2=x}} \right. \\ \\ \\ y=x-2 \\ y^2=(x-2)^2=x^2-4x+4 \\ \\ \\ y^2=x \\ x^2-4x+4=x \\ x^2-5x+4=0$


$a=1 \\ b=-5 \\ c=4 \\ \\ x=\frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}=\frac{5 \pm \sqrt{25 - 16}}{2}=\frac{5 \pm \sqrt{9}}{2}=\frac{5 \pm 3}{2} \\ \\ x=\frac{5+3}{2} \ \lor \ x=\frac{5-3}{2} \\ \\ x=\frac{8}{2} \ \lor \ x=\frac{2}{2} \\ \\ x=4 \ \lor \ x=1$


$y=4-2 \ \lor \ y=1-2 \\ y=2 \ \lor \ y=-1 \\ \\ \\ \left \{ {{x=4} \atop {y=2}} \right. \ \lor \ \left \{ {{x=1} \atop {y=-1}} \right.$

Answer 2

$y=x-2\\ y^2=x\\\\ y=y^2-2\\y^2-y-2=0\\ y^2+y-2y-2=0\\ y(y+1)-2(y+1)=0\\ (y-2)(y+1)=0\\ y=2 \vee y=-1\\\\ x=2^2 \vee x=(-1)^2\\ x=4 \vee x=1\\\\ \boxed{x=4,y=2 \vee x=1,y=-1}$