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

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

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

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

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solve for

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

x = \frac{z - y ^{2}}{2}, x = - \frac{z - y ^{2}}{2}

Solution steps

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

Switch sides

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

Move

y^{2}

to the right side

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

Divide both sides by

2

x^{2} = \frac{z - y ^{2}}{2}

For

x^{2} = f (a)

the solutions are

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

x = \frac{z - y ^{2}}{2}, x = - \frac{z - y ^{2}}{2}

- 11 = x^{2} - 36

x^{2} - 5 x - 24 = 0

3 x (x + 4) + (x + 4)^{2} = 0

6 y^{2} + 4 y

4 x + x^{2} = 5

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