What are the solutions to the equation 4y^2+1=0 ?

The solutions to the equation 4y^2+1=0 are y= 1/2 i,y=-1/2 i

Find the zeros of 4y^2+1=0

The zeros of 4y^2+1=0 are y= 1/2 i,y=-1/2 i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 y^{2} + 1 = 0

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

Solution steps

4 y^{2} + 1 = 0

Move

1

to the right side

4 y^{2} = - 1

Divide both sides by

4

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

For

x^{2} = f (a)

the solutions are

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

y = - \frac{1}{4}, y = - - \frac{1}{4}

- \frac{1}{4} = \frac{1}{2} i

- - \frac{1}{4} = - \frac{1}{2} i

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

\frac{( m ^{2} ) ^{6} m}{m ^{6}}

125 p^{3} + 8 q^{3}

x^{2} - 4 x \geq - 4

3 \cdot 6^{- 2}

- 9 + x = 4

What are the solutions to the equation 4y^2+1=0 ?
The solutions to the equation 4y^2+1=0 are y= 1/2 i,y=-1/2 i
Find the zeros of 4y^2+1=0
The zeros of 4y^2+1=0 are y= 1/2 i,y=-1/2 i