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

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

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

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

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solve for

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

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

Solution steps

z = x^{2} + 2 x y

Switch sides

x^{2} + 2 x y = z

Move

z

to the left side

x^{2} + 2 x y - z = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + 2 y x - z = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 y \pm ( 2 y ) ^{2} - 4 \cdot 1 \cdot ( - z )}{2 \cdot 1}

Simplify

(2 y)^{2} - 4 \cdot 1 \cdot (- z) : 2 y^{2} + z

x_{1, 2} = \frac{- 2 y \pm 2 y ^{2} + z}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 y + 2 y ^{2} + z}{2 \cdot 1}, x_{2} = \frac{- 2 y - 2 y ^{2} + z}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

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

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

s^{2} - 6 s + 11

s^{2} - 6 s + 13

x^{2} = 3 9^{2} + 16 0^{2}

- 0.3 x^{2} + 0.6 x = 22

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