What are the solutions to the equation (1/7)^2+y^2=1 ?

The solutions to the equation (1/7)^2+y^2=1 are y=(4sqrt(3))/7 ,y=-(4sqrt(3))/7

Find the zeros of (1/7)^2+y^2=1

The zeros of (1/7)^2+y^2=1 are y=(4sqrt(3))/7 ,y=-(4sqrt(3))/7

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(\frac{1}{7})^{2} + y^{2} = 1

y = \frac{4 3}{7}, y = - \frac{4 3}{7}

Decimal Notation

y = 0.98974 \dots, y = - 0.98974 \dots

Solution steps

(\frac{1}{7})^{2} + y^{2} = 1

Expand

(\frac{1}{7})^{2} + y^{2} : \frac{1}{49} + y^{2}

\frac{1}{49} + y^{2} = 1

Move

\frac{1}{49}

to the right side

y^{2} = 1 - \frac{1}{49}

Simplify

1 - \frac{1}{49} : \frac{48}{49}

For

x^{2} = f (a)

the solutions are

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

y = \frac{48}{49}, y = - \frac{48}{49}

\frac{48}{49} = \frac{4 3}{7}

- \frac{48}{49} = - \frac{4 3}{7}

y = \frac{4 3}{7}, y = - \frac{4 3}{7}

2.7 \cdot 1 0^{- 9} x^{2} - 16 \cdot 1 0^{- 5} x + 0.5 = 0

x, x^{2} - x y = 23

y^{2} - 8 y = 0

x, (x + k) (x^{2} + 3 x - 1) = x^{3} - x^{2} - 13 x + 4

m + 3 = m^{2} - 4 m - 3

What are the solutions to the equation (1/7)^2+y^2=1 ?
The solutions to the equation (1/7)^2+y^2=1 are y=(4sqrt(3))/7 ,y=-(4sqrt(3))/7
Find the zeros of (1/7)^2+y^2=1
The zeros of (1/7)^2+y^2=1 are y=(4sqrt(3))/7 ,y=-(4sqrt(3))/7