What are the solutions to the equation 2t^4+4t^3-4t^2=0 ?

The solutions to the equation 2t^4+4t^3-4t^2=0 are t=0,t=-1+sqrt(3),t=-1-sqrt(3)

Find the zeros of 2t^4+4t^3-4t^2=0

The zeros of 2t^4+4t^3-4t^2=0 are t=0,t=-1+sqrt(3),t=-1-sqrt(3)

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

t = 0, t = - 1 + 3, t = - 1 - 3

Solution steps

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

Factor

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

2 t^{2} (t^{2} + 2 t - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

t = 0 or t^{2} + 2 t - 2 = 0

Solve

t^{2} + 2 t - 2 = 0 : t = - 1 + 3, t = - 1 - 3

The solutions are

t = 0, t = - 1 + 3, t = - 1 - 3

16 x - 64 x^{3} = 0

x^{3} + 4 x^{2} + 6 x = 0

(x^{2} + 2 x) (x^{2} + 2 x - 7) = 8

- x^{4} + 7 x^{3} - 6 x^{2} = 0

x^{4} + 5 x^{2} - 14 = 0

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