What are the solutions to the equation 4x=x(2x^2-2) ?

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

Find the zeros of 4x=x(2x^2-2)

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

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4 x = x (2 x^{2} - 2)

x = 0, x = - 3, x = 3

Solution steps

4 x = x (2 x^{2} - 2)

Move

x (2 x^{2} - 2)

to the left side

4 x - x (2 x^{2} - 2) = 0

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 3 = 0 or x - 3 = 0

Solve

x + 3 = 0 : x = - 3

Solve

x - 3 = 0 : x = 3

The solutions are

x = 0, x = - 3, x = 3

(12 - 3 x) \cdot (2 x - 1) x = 0

x^{3} - 0.75 x - 0.25 = 0

- λ^{3} + 5 λ^{2} = 0

525 = 5 x^{3}

27 x^{4} - 7 x^{3} + 14 = 0

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