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

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

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

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

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solve for

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

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

Solution steps

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

Move

y^{2}

to the right side

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

Move

z^{2}

to the right side

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

For

x^{2} = f (a)

the solutions are

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

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

(x + 5)^{2} + (x - 2)^{2} = 37

a x^{2} + c = 0

4 x^{2} - 4 x + 3

2 x^{2} - 12 x + 16 = 0

x^{2} - 5 x + 6 = 0

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