What are the solutions to the equation 2x^4=432x ?

The solutions to the equation 2x^4=432x are x=0,x=6,x=-3+3sqrt(3)i,x=-3-3sqrt(3)i

Find the zeros of 2x^4=432x

The zeros of 2x^4=432x are x=0,x=6,x=-3+3sqrt(3)i,x=-3-3sqrt(3)i

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2 x^{4} = 432 x

x = 0, x = 6, x = - 3 + 33 i, x = - 3 - 33 i

Solution steps

2 x^{4} = 432 x

Move

432 x

to the left side

2 x^{4} - 432 x = 0

Factor

2 x^{4} - 432 x : 2 x (x - 6) (x^{2} + 6 x + 36)

2 x (x - 6) (x^{2} + 6 x + 36) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x - 6 = 0 or x^{2} + 6 x + 36 = 0

Solve

x - 6 = 0 : x = 6

Solve

x^{2} + 6 x + 36 = 0 : x = - 3 + 33 i, x = - 3 - 33 i

The solutions are

x = 0, x = 6, x = - 3 + 33 i, x = - 3 - 33 i

x^{5} - 80 x + 50 = 0

8 x^{5} - 4 x^{3} + 2 = 0

w^{3} - 2 w^{2} + w - 2 = 0.2

- 4 x^{5} + 32 x^{3} + 192 x = 0

- 2 x^{3} - x^{2} + 8 x - 5 = 0

What are the solutions to the equation 2x^4=432x ?
The solutions to the equation 2x^4=432x are x=0,x=6,x=-3+3sqrt(3)i,x=-3-3sqrt(3)i
Find the zeros of 2x^4=432x
The zeros of 2x^4=432x are x=0,x=6,x=-3+3sqrt(3)i,x=-3-3sqrt(3)i