What are the solutions to the equation solvefor y,x=y^2+2by-b^2 ?

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

Find the zeros of solvefor y,x=y^2+2by-b^2

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

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solve for

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

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

Solution steps

x = y^{2} + 2 b y - b^{2}

Switch sides

y^{2} + 2 b y - b^{2} = x

Move

x

to the left side

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

Solve with the quadratic formula

y_{1, 2} = \frac{- 2 b \pm ( 2 b ) ^{2} - 4 \cdot 1 \cdot ( - b ^{2} - x )}{2 \cdot 1}

Simplify

(2 b)^{2} - 4 \cdot 1 \cdot (- b^{2} - x) : 2 2 b^{2} + x

y_{1, 2} = \frac{- 2 b \pm 2 2 b ^{2} + x}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 2 b + 2 2 b ^{2} + x}{2 \cdot 1}, y_{2} = \frac{- 2 b - 2 2 b ^{2} + x}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

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

x (x + 10) = 144

x, 6 x^{2} - y^{2} = 0

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

\frac{1}{3} x^{2} - 2 x - 2 = 0

\frac{x ^{2}}{9} = \frac{1}{3}

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