What are the solutions to the equation solvefor x,z= 1/2 (x^2+y^2) ?

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

Find the zeros of solvefor x,z= 1/2 (x^2+y^2)

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

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solve for

x, z = \frac{1}{2} (x^{2} + y^{2})

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

Solution steps

z = \frac{1}{2} (x^{2} + y^{2})

Switch sides

\frac{1}{2} (x^{2} + y^{2}) = z

Multiply both sides by

2

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

Move

y^{2}

to the right side

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

For

x^{2} = f (a)

the solutions are

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

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

4 (- 3 k + 25)^{2} (1 - k) = (- 6 k + 29)^{2} (- 5 - k)

x^{2} + 8 x - 945 = 0

- x^{2} + 4 x = 7

x, x^{2} - 3 x y + 5 y^{2} - 28 x + 16 y - 30 = 0

- x^{2} + 64 = 0

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