What are the solutions to the equation solvefor y,x=y^2+4z^2-1 ?

The solutions to the equation solvefor y,x=y^2+4z^2-1 are y=sqrt(x-4z^2+1),y=-sqrt(x-4z^2+1)

Find the zeros of solvefor y,x=y^2+4z^2-1

The zeros of solvefor y,x=y^2+4z^2-1 are y=sqrt(x-4z^2+1),y=-sqrt(x-4z^2+1)

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solve for

y, x = y^{2} + 4 z^{2} - 1

y = x - 4 z^{2} + 1, y = - x - 4 z^{2} + 1

Solution steps

x = y^{2} + 4 z^{2} - 1

Switch sides

y^{2} + 4 z^{2} - 1 = x

Move

4 z^{2}

to the right side

y^{2} - 1 = x - 4 z^{2}

Move

1

to the right side

y^{2} = x - 4 z^{2} + 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

y = x - 4 z^{2} + 1, y = - x - 4 z^{2} + 1

x, x y (x + 2) (y - 4) = 3

x^{2} - 8 x - 6 = 4

- 3 x^{2} + 6 x + 21

x^{2} = 768

2 x^{2} - 42 = 0

What are the solutions to the equation solvefor y,x=y^2+4z^2-1 ?
The solutions to the equation solvefor y,x=y^2+4z^2-1 are y=sqrt(x-4z^2+1),y=-sqrt(x-4z^2+1)
Find the zeros of solvefor y,x=y^2+4z^2-1
The zeros of solvefor y,x=y^2+4z^2-1 are y=sqrt(x-4z^2+1),y=-sqrt(x-4z^2+1)