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

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

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

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

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solve for

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

x = z - 2 y^{2} + 1, x = - z - 2 y^{2} + 1

Solution steps

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

Switch sides

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

Move

2 y^{2}

to the right side

(x - 1)^{2} = z - 2 y^{2}

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

x - 1 = z - 2 y^{2} : x = z - 2 y^{2} + 1

Solve

x - 1 = - z - 2 y^{2} : x = - z - 2 y^{2} + 1

The solutions to the quadratic equation are:

x = z - 2 y^{2} + 1, x = - z - 2 y^{2} + 1

3 X^{2} + 9 X - 2 = 0

4 (\frac{2}{3}) x^{2} - 5 x + 4 = 9

- 3 x (x + 2) = 0

λ^{2} + 12 λ = 59

\frac{6.43 - x}{500} = 0.0061 (x - 0.994)^{2}

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