What are the solutions to the equation 3y=4y^3 ?

The solutions to the equation 3y=4y^3 are y=0,y=-(sqrt(3))/2 ,y=(sqrt(3))/2

Find the zeros of 3y=4y^3

The zeros of 3y=4y^3 are y=0,y=-(sqrt(3))/2 ,y=(sqrt(3))/2

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3 y = 4 y^{3}

y = 0, y = - \frac{3}{2}, y = \frac{3}{2}

Decimal Notation

y = 0, y = - 0.86602 \dots, y = 0.86602 \dots

Solution steps

3 y = 4 y^{3}

Move

4 y^{3}

to the left side

3 y - 4 y^{3} = 0

Factor

3 y - 4 y^{3} : - y (2 y + 3) (2 y - 3)

- y (2 y + 3) (2 y - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0 or 2 y + 3 = 0 or 2 y - 3 = 0

Solve

2 y + 3 = 0 : y = - \frac{3}{2}

Solve

2 y - 3 = 0 : y = \frac{3}{2}

The solutions are

y = 0, y = - \frac{3}{2}, y = \frac{3}{2}

30 π r^{2} + 4 π r^{3} - 4500 = 0

2 x^{3} - 9 x^{2} + 7 x + 6 = 0

(3 x^{2} + 7 x + 2) (3 x - a) = 0

x^{4} - 4 x^{3} + 3 x^{2} = 0

6 x^{3} = 48

What are the solutions to the equation 3y=4y^3 ?
The solutions to the equation 3y=4y^3 are y=0,y=-(sqrt(3))/2 ,y=(sqrt(3))/2
Find the zeros of 3y=4y^3
The zeros of 3y=4y^3 are y=0,y=-(sqrt(3))/2 ,y=(sqrt(3))/2