What are the solutions to the equation (1/8)^2+y^2=1 ?

The solutions to the equation (1/8)^2+y^2=1 are y=(3sqrt(7))/8 ,y=-(3sqrt(7))/8

Find the zeros of (1/8)^2+y^2=1

The zeros of (1/8)^2+y^2=1 are y=(3sqrt(7))/8 ,y=-(3sqrt(7))/8

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(\frac{1}{8})^{2} + y^{2} = 1

y = \frac{3 7}{8}, y = - \frac{3 7}{8}

Decimal Notation

y = 0.99215 \dots, y = - 0.99215 \dots

Solution steps

(\frac{1}{8})^{2} + y^{2} = 1

Expand

(\frac{1}{8})^{2} + y^{2} : \frac{1}{64} + y^{2}

\frac{1}{64} + y^{2} = 1

Move

\frac{1}{64}

to the right side

y^{2} = 1 - \frac{1}{64}

Simplify

1 - \frac{1}{64} : \frac{63}{64}

For

x^{2} = f (a)

the solutions are

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

y = \frac{63}{64}, y = - \frac{63}{64}

\frac{63}{64} = \frac{3 7}{8}

- \frac{63}{64} = - \frac{3 7}{8}

y = \frac{3 7}{8}, y = - \frac{3 7}{8}

\frac{x}{6} + \frac{x}{12} + \frac{x}{7} + 5 + \frac{x}{2} + 4 = x

6 x > 12

2 n^{2} + 9 n + 10

\frac{i \frac{1}{2}}{i \frac{3}{2}}

- 2 + \frac{4}{5} x = 6

What are the solutions to the equation (1/8)^2+y^2=1 ?
The solutions to the equation (1/8)^2+y^2=1 are y=(3sqrt(7))/8 ,y=-(3sqrt(7))/8
Find the zeros of (1/8)^2+y^2=1
The zeros of (1/8)^2+y^2=1 are y=(3sqrt(7))/8 ,y=-(3sqrt(7))/8