What is the answer to solvefor y,x=sqrt(y^2+2y) ?

The answer to solvefor y,x=sqrt(y^2+2y) is y=-1+sqrt(x^2+1) {x>= 0},y=-sqrt(x^2+1)-1 {x>= 0}

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solvefor y,x=sqrt(y^2+2y)

solve for

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

y = - 1 + x^{2} + 1 {x \geq 0}, y = - x^{2} + 1 - 1 {x \geq 0}

Solution steps

x = y^{2} + 2 y

Square both sides:

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

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

Solve

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

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

Verify Solutions:

y = - 1 + x^{2} + 1 {x \geq 0}, y = - x^{2} + 1 - 1 {x \geq 0}

The solutions are

y = - 1 + x^{2} + 1 {x \geq 0}, y = - x^{2} + 1 - 1 {x \geq 0}

(1 - x)^{0.5} = \frac{3}{2}

(x) = x^{\frac{1}{3}} (x + 4)

p^{\frac{- 5}{2}} = 2^{- 5}

7 - 3 p = p

x^{2} + x^{2} = 20

What is the answer to solvefor y,x=sqrt(y^2+2y) ?
The answer to solvefor y,x=sqrt(y^2+2y) is y=-1+sqrt(x^2+1) {x>= 0},y=-sqrt(x^2+1)-1 {x>= 0}