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

The solutions to the equation x(x+1)^2=(x-1)^3 are x= 1/5-i 2/5 ,x= 1/5+i 2/5

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

The zeros of x(x+1)^2=(x-1)^3 are x= 1/5-i 2/5 ,x= 1/5+i 2/5

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

x = \frac{1}{5} - i \frac{2}{5}, x = \frac{1}{5} + i \frac{2}{5}

Solution steps

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

Expand

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

Expand

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

x^{3} + 2 x^{2} + x = x^{3} - 3 x^{2} + 3 x - 1

Switch sides

x^{3} - 3 x^{2} + 3 x - 1 = x^{3} + 2 x^{2} + x

Move

x

to the left side

x^{3} - 3 x^{2} + 2 x - 1 = x^{3} + 2 x^{2}

Move

2 x^{2}

to the left side

x^{3} - 5 x^{2} + 2 x - 1 = x^{3}

Move

x^{3}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 5 ) ( - 1 )}{2 ( - 5 )}

Simplify

2^{2} - 4 (- 5) (- 1) : 4 i

x_{1, 2} = \frac{- 2 \pm 4 i}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- 2 + 4 i}{2 ( - 5 )}, x_{2} = \frac{- 2 - 4 i}{2 ( - 5 )}

x = \frac{- 2 + 4 i}{2 ( - 5 )} : \frac{1}{5} - i \frac{2}{5}

x = \frac{- 2 - 4 i}{2 ( - 5 )} : \frac{1}{5} + i \frac{2}{5}

The solutions to the quadratic equation are:

x = \frac{1}{5} - i \frac{2}{5}, x = \frac{1}{5} + i \frac{2}{5}

x^{2} + 4 x + 12 = - 2 x + 4

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

4 x^{2} + 12 x + 14

- x^{2} + x + 6

0 = y^{2} - 65

What are the solutions to the equation x(x+1)^2=(x-1)^3 ?
The solutions to the equation x(x+1)^2=(x-1)^3 are x= 1/5-i 2/5 ,x= 1/5+i 2/5
Find the zeros of x(x+1)^2=(x-1)^3
The zeros of x(x+1)^2=(x-1)^3 are x= 1/5-i 2/5 ,x= 1/5+i 2/5