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

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

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

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

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

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

Solution steps

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

Switch sides

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

Move

2

to the left side

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

Move

4 x

to the left side

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

Move

3 x^{2}

to the left side

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

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

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

Factor

x^{3} - 3 x^{2} + 3 x - 2 : (x - 2) (x^{2} - x + 1)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

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

The solutions are

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

2.5 x^{4} + 3 x^{3} - 2.6 x^{2} - 5.1 x - 5.6

x^{5} = 310

3 (4 x^{12}) = 10

x^{3} + 3 x^{2} - 14 x - 20 = 0

x^{3} - 16 x^{2} + 77 x - 98 = 0

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