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}

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}

a^{2} - a

\frac{c ^{4} d}{c d}

y + 7 + 2 y = - 14 + 3 y + 21

4 x^{2} - 20 x = - 24

3 x^{2} + 10 x + 3 = 0

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