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

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

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

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

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solve for

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

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

Solution steps

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

Switch sides

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

Move

4 z^{2}

to the right side

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

For

x^{2} = f (a)

the solutions are

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

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

- 5 x^{2} + 10 x = 5

x^{2} + 4 x - 4 = 0

x^{2} + 4 x + 52 = 0

x, x^{2} + y^{2} = 2 a x

0 = x^{2} - 5 x - 36

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