What are the solutions to the equation (-1/3)^2+y^2=1 ?

The solutions to the equation (-1/3)^2+y^2=1 are y=(2sqrt(2))/3 ,y=-(2sqrt(2))/3

Find the zeros of (-1/3)^2+y^2=1

The zeros of (-1/3)^2+y^2=1 are y=(2sqrt(2))/3 ,y=-(2sqrt(2))/3

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(- \frac{1}{3})^{2} + y^{2} = 1

y = \frac{2 2}{3}, y = - \frac{2 2}{3}

Decimal Notation

y = 0.94280 \dots, y = - 0.94280 \dots

Solution steps

(- \frac{1}{3})^{2} + y^{2} = 1

Expand

(- \frac{1}{3})^{2} + y^{2} : \frac{1}{9} + y^{2}

\frac{1}{9} + y^{2} = 1

Move

\frac{1}{9}

to the right side

y^{2} = 1 - \frac{1}{9}

Simplify

1 - \frac{1}{9} : \frac{8}{9}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

y = \frac{8}{9}, y = - \frac{8}{9}

\frac{8}{9} = \frac{2 2}{3}

- \frac{8}{9} = - \frac{2 2}{3}

y = \frac{2 2}{3}, y = - \frac{2 2}{3}

- 4.9 t^{2} + 73.92 = \frac{1}{2} \cdot 73.92

z^{2} + 6 z + 10

2 v^{2} + 5 v = 7

2 x^{2} - 12 = - 15

(- x + 4)^{2} = 40

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