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

The solutions to the equation 2(1/3)^2+y^2=1 are y=(sqrt(7))/3 ,y=-(sqrt(7))/3

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

The zeros of 2(1/3)^2+y^2=1 are y=(sqrt(7))/3 ,y=-(sqrt(7))/3

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

y = \frac{7}{3}, y = - \frac{7}{3}

Decimal Notation

y = 0.88191 \dots, y = - 0.88191 \dots

Solution steps

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

Expand

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

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

Move

\frac{2}{9}

to the right side

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

Simplify

1 - \frac{2}{9} : \frac{7}{9}

For

x^{2} = f (a)

the solutions are

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

y = \frac{7}{9}, y = - \frac{7}{9}

\frac{7}{9} = \frac{7}{3}

- \frac{7}{9} = - \frac{7}{3}

y = \frac{7}{3}, y = - \frac{7}{3}

- x^{2} - 14 x - 13 = 0

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

x^{2} - 3 x - 4 = - 3 x

5 d^{2} - 8 d + 1 = 0

1 3^{2} + 9^{2} = c^{2}

What are the solutions to the equation 2(1/3)^2+y^2=1 ?
The solutions to the equation 2(1/3)^2+y^2=1 are y=(sqrt(7))/3 ,y=-(sqrt(7))/3
Find the zeros of 2(1/3)^2+y^2=1
The zeros of 2(1/3)^2+y^2=1 are y=(sqrt(7))/3 ,y=-(sqrt(7))/3