What are the solutions to the equation solvefor y,y^2=15x^3+2xy ?

The solutions to the equation solvefor y,y^2=15x^3+2xy are y=x(sqrt(15x+1)+1),y=x(-sqrt(15x+1)+1)

Find the zeros of solvefor y,y^2=15x^3+2xy

The zeros of solvefor y,y^2=15x^3+2xy are y=x(sqrt(15x+1)+1),y=x(-sqrt(15x+1)+1)

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solve for

y, y^{2} = 15 x^{3} + 2 x y

y = x (15 x + 1 + 1), y = x (- 15 x + 1 + 1)

Solution steps

y^{2} = 15 x^{3} + 2 x y

Move

2 x y

to the left side

y^{2} - 2 x y = 15 x^{3}

Move

15 x^{3}

to the left side

y^{2} - 2 x y - 15 x^{3} = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 2 x ) \pm ( - 2 x ) ^{2} - 4 \cdot 1 \cdot ( - 15 x ^{3} )}{2 \cdot 1}

Simplify

(- 2 x)^{2} - 4 \cdot 1 \cdot (- 15 x^{3}) : 2 x 1 + 15 x

y_{1, 2} = \frac{- ( - 2 x ) \pm 2 x 1 + 15 x}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 2 x ) + 2 x 1 + 15 x}{2 \cdot 1}, y_{2} = \frac{- ( - 2 x ) - 2 x 1 + 15 x}{2 \cdot 1}

y = \frac{- ( - 2 x ) + 2 x 1 + 15 x}{2 \cdot 1} : x (15 x + 1 + 1)

y = \frac{- ( - 2 x ) - 2 x 1 + 15 x}{2 \cdot 1} : x (- 15 x + 1 + 1)

The solutions to the quadratic equation are:

y = x (15 x + 1 + 1), y = x (- 15 x + 1 + 1)

(1 - x) (- 3 - x) - 2 = 0

- 2 x^{2} - 16 x - 35 = 0

(2 t - 1) t = 66

(x + 2)^{2} = 2 (x + 6)

t^{2} - 2 t + 0.4082 = 0

What are the solutions to the equation solvefor y,y^2=15x^3+2xy ?
The solutions to the equation solvefor y,y^2=15x^3+2xy are y=x(sqrt(15x+1)+1),y=x(-sqrt(15x+1)+1)
Find the zeros of solvefor y,y^2=15x^3+2xy
The zeros of solvefor y,y^2=15x^3+2xy are y=x(sqrt(15x+1)+1),y=x(-sqrt(15x+1)+1)