What are the solutions to the equation 6-3(x+1)=-4x^2(1-x) ?

The solutions to the equation 6-3(x+1)=-4x^2(1-x) are x=1,x=(sqrt(3))/2 i,x=-(sqrt(3))/2 i

Find the zeros of 6-3(x+1)=-4x^2(1-x)

The zeros of 6-3(x+1)=-4x^2(1-x) are x=1,x=(sqrt(3))/2 i,x=-(sqrt(3))/2 i

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6 - 3 (x + 1) = - 4 x^{2} (1 - x)

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

Solution steps

6 - 3 (x + 1) = - 4 x^{2} (1 - x)

Expand

6 - 3 (x + 1) : - 3 x + 3

Expand

- 4 x^{2} (1 - x) : - 4 x^{2} + 4 x^{3}

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

Switch sides

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

Move

3

to the left side

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

Move

3 x

to the left side

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

Write in the standard form

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

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

Factor

4 x^{3} - 4 x^{2} + 3 x - 3 : (x - 1) (4 x^{2} + 3)

(x - 1) (4 x^{2} + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or 4 x^{2} + 3 = 0

Solve

x - 1 = 0 : x = 1

Solve

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

The solutions are

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

3 x^{4} - 3 x^{3} - 1 x^{2} = 0

z \cdot z^{3} + 15 = 16.66

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

x^{4} - (4 \cdot x) + 5 = 0

y^{5} + y^{3} + y = 3

What are the solutions to the equation 6-3(x+1)=-4x^2(1-x) ?
The solutions to the equation 6-3(x+1)=-4x^2(1-x) are x=1,x=(sqrt(3))/2 i,x=-(sqrt(3))/2 i
Find the zeros of 6-3(x+1)=-4x^2(1-x)
The zeros of 6-3(x+1)=-4x^2(1-x) are x=1,x=(sqrt(3))/2 i,x=-(sqrt(3))/2 i