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Nat2105 [25]
3 years ago
11

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

Mathematics
2 answers:
omeli [17]3 years ago
8 0
\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.
IRISSAK [1]3 years ago
6 0
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}


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