What are the solutions to the equation 4(1/3)^2+y^2=4 ?

The solutions to the equation 4(1/3)^2+y^2=4 are y=(4sqrt(2))/3 ,y=-(4sqrt(2))/3

Find the zeros of 4(1/3)^2+y^2=4

The zeros of 4(1/3)^2+y^2=4 are y=(4sqrt(2))/3 ,y=-(4sqrt(2))/3

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4 (\frac{1}{3})^{2} + y^{2} = 4

y = \frac{4 2}{3}, y = - \frac{4 2}{3}

Decimal Notation

y = 1.88561 \dots, y = - 1.88561 \dots

Solution steps

4 (\frac{1}{3})^{2} + y^{2} = 4

Expand

4 (\frac{1}{3})^{2} + y^{2} : \frac{4}{9} + y^{2}

\frac{4}{9} + y^{2} = 4

Move

\frac{4}{9}

to the right side

y^{2} = 4 - \frac{4}{9}

Simplify

4 - \frac{4}{9} : \frac{32}{9}

For

x^{2} = f (a)

the solutions are

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

y = \frac{32}{9}, y = - \frac{32}{9}

\frac{32}{9} = \frac{4 2}{3}

- \frac{32}{9} = - \frac{4 2}{3}

y = \frac{4 2}{3}, y = - \frac{4 2}{3}

1 = x^{2} - 3

1 = x^{2} - 4

5 a^{2} - 12 a + 3 = 0

x^{2} + 6 x - 55 = 0

2 x^{2} - 7 = 11

What are the solutions to the equation 4(1/3)^2+y^2=4 ?
The solutions to the equation 4(1/3)^2+y^2=4 are y=(4sqrt(2))/3 ,y=-(4sqrt(2))/3
Find the zeros of 4(1/3)^2+y^2=4
The zeros of 4(1/3)^2+y^2=4 are y=(4sqrt(2))/3 ,y=-(4sqrt(2))/3