What are the solutions to the equation 1/4+y^2=1 ?

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

Find the zeros of 1/4+y^2=1

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

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\frac{1}{4} + 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}{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}

y, x^{3} + x y - y^{2} = 0

7 r^{2} - 16 k + 3 = 3

- 24 = - 5 b - b^{2}

x, x^{2} = y^{4}

3 (x + 12) (x - 12) = 0

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