What are the solutions to the equation 36x^3+24x^2-4x=0 ?

The solutions to the equation 36x^3+24x^2-4x=0 are x=0,x=(-1+sqrt(2))/3 ,x=-(1+sqrt(2))/3

Find the zeros of 36x^3+24x^2-4x=0

The zeros of 36x^3+24x^2-4x=0 are x=0,x=(-1+sqrt(2))/3 ,x=-(1+sqrt(2))/3

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36 x^{3} + 24 x^{2} - 4 x = 0

x = 0, x = \frac{- 1 + 2}{3}, x = - \frac{1 + 2}{3}

Decimal Notation

x = 0, x = 0.13807 \dots, x = - 0.80473 \dots

Solution steps

36 x^{3} + 24 x^{2} - 4 x = 0

Factor

36 x^{3} + 24 x^{2} - 4 x : 4 x (9 x^{2} + 6 x - 1)

4 x (9 x^{2} + 6 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 9 x^{2} + 6 x - 1 = 0

Solve

9 x^{2} + 6 x - 1 = 0 : x = \frac{- 1 + 2}{3}, x = - \frac{1 + 2}{3}

The solutions are

x = 0, x = \frac{- 1 + 2}{3}, x = - \frac{1 + 2}{3}

x^{3} + x^{2} - 10 x + 8 = 0

x^{3} = 15

k^{3} - 4 k^{2} + k + 6 = 0

x^{3} - 10 x^{2} + 31 x - 30 = 0

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

What are the solutions to the equation 36x^3+24x^2-4x=0 ?
The solutions to the equation 36x^3+24x^2-4x=0 are x=0,x=(-1+sqrt(2))/3 ,x=-(1+sqrt(2))/3
Find the zeros of 36x^3+24x^2-4x=0
The zeros of 36x^3+24x^2-4x=0 are x=0,x=(-1+sqrt(2))/3 ,x=-(1+sqrt(2))/3