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

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

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

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

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solve for

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

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

Solution steps

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

Switch sides

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

Move

y^{2}

to the right side

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

For

x^{2} = f (a)

the solutions are

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

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

(x + 6)^{2} = 25

\frac{2 x ^{2}}{5} + \frac{5 x}{4} = 10

a, c^{2} = a^{2} + b^{2}

x, 5 x^{2} - 2 x y + 2 y^{2} - 2 y - 71 = 0

6 x^{2} + 4 x = 0

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