- Thread starter Ilikebugs
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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(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.

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.

- Joined
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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.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