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

The solutions to the equation (x-4)^3=x(x^2+2x) are x= 12/7-i(4sqrt(5))/7 ,x= 12/7+i(4sqrt(5))/7

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

The zeros of (x-4)^3=x(x^2+2x) are x= 12/7-i(4sqrt(5))/7 ,x= 12/7+i(4sqrt(5))/7

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

x = \frac{12}{7} - i \frac{4 5}{7}, x = \frac{12}{7} + i \frac{4 5}{7}

Solution steps

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

Expand

(x - 4)^{3} : x^{3} - 12 x^{2} + 48 x - 64

Expand

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

x^{3} - 12 x^{2} + 48 x - 64 = x^{3} + 2 x^{2}

Move

2 x^{2}

to the left side

x^{3} - 14 x^{2} + 48 x - 64 = x^{3}

Move

x^{3}

to the left side

- 14 x^{2} + 48 x - 64 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 14 ) ( - 64 )}{2 ( - 14 )}

Simplify

4 8^{2} - 4 (- 14) (- 64) : 165 i

x_{1, 2} = \frac{- 48 \pm 16 5 i}{2 ( - 14 )}

Separate the solutions

x_{1} = \frac{- 48 + 16 5 i}{2 ( - 14 )}, x_{2} = \frac{- 48 - 16 5 i}{2 ( - 14 )}

x = \frac{- 48 + 16 5 i}{2 ( - 14 )} : \frac{12}{7} - i \frac{4 5}{7}

x = \frac{- 48 - 16 5 i}{2 ( - 14 )} : \frac{12}{7} + i \frac{4 5}{7}

The solutions to the quadratic equation are:

x = \frac{12}{7} - i \frac{4 5}{7}, x = \frac{12}{7} + i \frac{4 5}{7}

m, (- 3 m)^{2} = 4 p - 8

x, \frac{r}{f} (x) = 2 x^{2} - 2 x - 12

a (x^{2} + 1) = x (a^{2} + 1)

x, f (x^{2} + y^{2} + z^{2} z^{2} - 2 x y) = 0

3 = xx + xx + xx

What are the solutions to the equation (x-4)^3=x(x^2+2x) ?
The solutions to the equation (x-4)^3=x(x^2+2x) are x= 12/7-i(4sqrt(5))/7 ,x= 12/7+i(4sqrt(5))/7
Find the zeros of (x-4)^3=x(x^2+2x)
The zeros of (x-4)^3=x(x^2+2x) are x= 12/7-i(4sqrt(5))/7 ,x= 12/7+i(4sqrt(5))/7