What are the solutions to the equation (15/17)^2+y^2=1 ?

The solutions to the equation (15/17)^2+y^2=1 are y= 8/17 ,y=-8/17

Find the zeros of (15/17)^2+y^2=1

The zeros of (15/17)^2+y^2=1 are y= 8/17 ,y=-8/17

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(\frac{15}{17})^{2} + y^{2} = 1

y = \frac{8}{17}, y = - \frac{8}{17}

Solution steps

(\frac{15}{17})^{2} + y^{2} = 1

Expand

(\frac{15}{17})^{2} + y^{2} : \frac{225}{289} + y^{2}

\frac{225}{289} + y^{2} = 1

Move

\frac{225}{289}

to the right side

y^{2} = 1 - \frac{225}{289}

Simplify

1 - \frac{225}{289} : \frac{64}{289}

For

x^{2} = f (a)

the solutions are

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

y = \frac{64}{289}, y = - \frac{64}{289}

\frac{64}{289} = \frac{8}{17}

- \frac{64}{289} = - \frac{8}{17}

y = \frac{8}{17}, y = - \frac{8}{17}

3 c^{2} - 10 c + 8

- 2 (x + 2) < 14

11 s^{2} - 26 s - 61 = 0

x^{2} + 16 x + 49 = 0

9 + x^{2} - 6

What are the solutions to the equation (15/17)^2+y^2=1 ?
The solutions to the equation (15/17)^2+y^2=1 are y= 8/17 ,y=-8/17
Find the zeros of (15/17)^2+y^2=1
The zeros of (15/17)^2+y^2=1 are y= 8/17 ,y=-8/17