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

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

Find the zeros of 2y^2=1

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

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2 y^{2} = 1

y = \frac{1}{2}, y = - \frac{1}{2}

Solution steps

2 y^{2} = 1

Divide both sides by

2

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

For

x^{2} = f (a)

the solutions are

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

y = \frac{1}{2}, y = - \frac{1}{2}

\frac{1}{2} = \frac{1}{2}

- \frac{1}{2} = - \frac{1}{2}

y = \frac{1}{2}, y = - \frac{1}{2}

(x + 7)^{2} = - 25

64 x^{2} - 4 = 0

3 (x - 1)^{2} = - 27

- 3 x^{2} + x = 13

1.5 x^{2} + 24 = 0

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