Point of intersection of circle (x-2)^2+(y-3)^2=6, parabola y=-2x^2+14x-1

Ilikebugs

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Point of intersection of circle and parabola

(x-2)^2+(y-3)^2=6 and y=-2x^2+14x-1

These are the two equations, but I dont know how to start finding all the point of intersections.
 
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Jomo

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(x-2)^2+(y-3)^2=6 and y=-2x^2+14x-1 are the two equations, but I dont know how to start finding all the point of intersections.
You should have learned two standard ways to find the points of intersections of two curves. Have you been going to class. I'll get you started by telling the two typical methods

1) You solve both equations for y (or x) and set the equations equal to one another and then solve for x (or y)

2) Substitute the expression for y (or x) into the other equation and then solve for x (or y).

Try one of these and show us your work.
 

Ilikebugs

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x^2 - 4x + y^2 - 6y = -7 [1]

(-2x^2 + 14x - 1) = y [2]



x^2 - 4x + (-2x^2 + 14x - 1)^2 - 6(-2x^2 + 14x - 1) = -7
4x^4-56x^3+213x^2-116x+14=0

I never learned how to solve quartic equations tho
 

Jomo

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Joined
Dec 30, 2014
Messages
3,050
x^2 - 4x + y^2 - 6y = -7 [1]

(-2x^2 + 14x - 1) = y [2]



x^2 - 4x + (-2x^2 + 14x - 1)^2 - 6(-2x^2 + 14x - 1) = -7
4x^4-56x^3+213x^2-116x+14=0

I never learned how to solve quartic equations tho
At the level of beginning algebra (which is where you posted your problem) you should be able to see the roots easily by trying +/- 1 and/or +/- 2 but these do not work. In fact the roots are irrational. This leads me to think that maybe you copied down the wrong problem or your textbook/teacher made a mistake.
 
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