What are the solutions to the equation (9/41)^2+y^2=1 ?

The solutions to the equation (9/41)^2+y^2=1 are y= 40/41 ,y=-40/41

Find the zeros of (9/41)^2+y^2=1

The zeros of (9/41)^2+y^2=1 are y= 40/41 ,y=-40/41

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(\frac{9}{41})^{2} + y^{2} = 1

y = \frac{40}{41}, y = - \frac{40}{41}

Decimal Notation

y = 0.97560 \dots, y = - 0.97560 \dots

Solution steps

(\frac{9}{41})^{2} + y^{2} = 1

Expand

(\frac{9}{41})^{2} + y^{2} : \frac{81}{1681} + y^{2}

\frac{81}{1681} + y^{2} = 1

Move

\frac{81}{1681}

to the right side

y^{2} = 1 - \frac{81}{1681}

Simplify

1 - \frac{81}{1681} : \frac{1600}{1681}

For

x^{2} = f (a)

the solutions are

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

y = \frac{1600}{1681}, y = - \frac{1600}{1681}

\frac{1600}{1681} = \frac{40}{41}

- \frac{1600}{1681} = - \frac{40}{41}

y = \frac{40}{41}, y = - \frac{40}{41}

1 6^{2} = x^{2} + 6^{2}

13 x^{2} - 3 x = 4 x^{2} + 21

z, g = k (z - 0.8) (z + 0.7)

0 = (x - 1)^{2} + 2

x^{2} + 24 x + 19 = - 25

What are the solutions to the equation (9/41)^2+y^2=1 ?
The solutions to the equation (9/41)^2+y^2=1 are y= 40/41 ,y=-40/41
Find the zeros of (9/41)^2+y^2=1
The zeros of (9/41)^2+y^2=1 are y= 40/41 ,y=-40/41