What are the solutions to the equation 3y^2+12y^4=(3y^2) ?

The solutions to the equation 3y^2+12y^4=(3y^2) are y=0

Find the zeros of 3y^2+12y^4=(3y^2)

The zeros of 3y^2+12y^4=(3y^2) are y=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 y^{2} + 12 y^{4} = (3 y^{2})

y = 0

Solution steps

3 y^{2} + 12 y^{4} = (3 y^{2})

Refine

3 y^{2} + 12 y^{4} = 3 y^{2}

Move

3 y^{2}

to the left side

12 y^{4} = 0

Divide both sides by

12

y^{4} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0

The solution is

y = 0

x^{4} - 21 x^{2} - 100 = 0

3 x (x - 1) (x - 2) = 0

0 = x^{3} - 9 x

6 x^{3} - 31 x^{2} + 3 x + 10 = 0

y^{3} - 27 = 9 y^{2} - 27 y

What are the solutions to the equation 3y^2+12y^4=(3y^2) ?
The solutions to the equation 3y^2+12y^4=(3y^2) are y=0
Find the zeros of 3y^2+12y^4=(3y^2)
The zeros of 3y^2+12y^4=(3y^2) are y=0