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

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

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

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

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

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

Decimal Notation

y = 0.86602 \dots, y = - 0.86602 \dots

Solution steps

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

Expand

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

\frac{1}{4} + y^{2} = 1

Move

\frac{1}{4}

to the right side

y^{2} = 1 - \frac{1}{4}

Simplify

1 - \frac{1}{4} : \frac{3}{4}

For

x^{2} = f (a)

the solutions are

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

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

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

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

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

x^{2} + 10 x + \frac{8}{3} = 0

5 x^{2} + 9 x + 3 = 0

- x^{2} - 27 = 0

x^{2} + 1 1^{2} = 1 4^{2}

49 m^{2} + 9 = 109

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