What are the solutions to the equation (-(sqrt(5))/3)^2+y^2=1 ?

The solutions to the equation (-(sqrt(5))/3)^2+y^2=1 are y= 2/3 ,y=-2/3

Find the zeros of (-(sqrt(5))/3)^2+y^2=1

The zeros of (-(sqrt(5))/3)^2+y^2=1 are y= 2/3 ,y=-2/3

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(- \frac{5}{3})^{2} + y^{2} = 1

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

Decimal Notation

y = 0.66666 \dots, y = - 0.66666 \dots

Solution steps

(- \frac{5}{3})^{2} + y^{2} = 1

Expand

(- \frac{5}{3})^{2} + y^{2} : \frac{5}{9} + y^{2}

\frac{5}{9} + y^{2} = 1

Move

\frac{5}{9}

to the right side

y^{2} = 1 - \frac{5}{9}

Simplify

1 - \frac{5}{9} : \frac{4}{9}

For

x^{2} = f (a)

the solutions are

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

y = \frac{4}{9}, y = - \frac{4}{9}

\frac{4}{9} = \frac{2}{3}

- \frac{4}{9} = - \frac{2}{3}

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

- x^{2} + 4 x - 20

x, x^{2} + y^{2} - 2 x - 6 y + 10 = 0

3 n^{2} + 8 n - 3 = 0

3^{2} + x^{2} = 1 0^{2}

- 4.9 x^{2} + 26.19 x = 0

What are the solutions to the equation (-(sqrt(5))/3)^2+y^2=1 ?
The solutions to the equation (-(sqrt(5))/3)^2+y^2=1 are y= 2/3 ,y=-2/3
Find the zeros of (-(sqrt(5))/3)^2+y^2=1
The zeros of (-(sqrt(5))/3)^2+y^2=1 are y= 2/3 ,y=-2/3