What are the solutions to the equation 1^2+y^2=1+8y ?

The solutions to the equation 1^2+y^2=1+8y are y=8,y=0

Find the zeros of 1^2+y^2=1+8y

The zeros of 1^2+y^2=1+8y are y=8,y=0

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1^{2} + y^{2} = 1 + 8 y

y = 8, y = 0

Solution steps

1^{2} + y^{2} = 1 + 8 y

Expand

1^{2} + y^{2} : 1 + y^{2}

1 + y^{2} = 1 + 8 y

Move

8 y

to the left side

1 + y^{2} - 8 y = 1

Move

1

to the left side

y^{2} - 8 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 0 = 8

y_{1, 2} = \frac{- ( - 8 ) \pm 8}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 8}{2 \cdot 1}, y_{2} = \frac{- ( - 8 ) - 8}{2 \cdot 1}

y = \frac{- ( - 8 ) + 8}{2 \cdot 1} : 8

y = \frac{- ( - 8 ) - 8}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

y = 8, y = 0

x^{2} + 0.2 x + 9.01 = 0

x, 2 x^{2} - 4 x + 1 = 0

192 = - 16 t^{2} + 112 t

- 2 x^{2} = 8 x

(\frac{5}{13})^{2} + y^{2} = 1

What are the solutions to the equation 1^2+y^2=1+8y ?
The solutions to the equation 1^2+y^2=1+8y are y=8,y=0
Find the zeros of 1^2+y^2=1+8y
The zeros of 1^2+y^2=1+8y are y=8,y=0