What are the solutions to the equation x(x+5)=(x^2+5x)^2 ?

The solutions to the equation x(x+5)=(x^2+5x)^2 are x=0,x=-5,x=(-5+sqrt(29))/2 ,x=(-5-sqrt(29))/2

Find the zeros of x(x+5)=(x^2+5x)^2

The zeros of x(x+5)=(x^2+5x)^2 are x=0,x=-5,x=(-5+sqrt(29))/2 ,x=(-5-sqrt(29))/2

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x (x + 5) = (x^{2} + 5 x)^{2}

x = 0, x = - 5, x = \frac{- 5 + 29}{2}, x = \frac{- 5 - 29}{2}

Decimal Notation

x = 0, x = - 5, x = 0.19258 \dots, x = - 5.19258 \dots

Solution steps

x (x + 5) = (x^{2} + 5 x)^{2}

Expand

x (x + 5) : x^{2} + 5 x

Expand

(x^{2} + 5 x)^{2} : x^{4} + 10 x^{3} + 25 x^{2}

x^{2} + 5 x = x^{4} + 10 x^{3} + 25 x^{2}

Switch sides

x^{4} + 10 x^{3} + 25 x^{2} = x^{2} + 5 x

Move

5 x

to the left side

x^{4} + 10 x^{3} + 25 x^{2} - 5 x = x^{2}

Move

x^{2}

to the left side

x^{4} + 10 x^{3} + 24 x^{2} - 5 x = 0

Factor

x^{4} + 10 x^{3} + 24 x^{2} - 5 x : x (x + 5) (x^{2} + 5 x - 1)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 5 = 0 : x = - 5

Solve

x^{2} + 5 x - 1 = 0 : x = \frac{- 5 + 29}{2}, x = \frac{- 5 - 29}{2}

The solutions are

x = 0, x = - 5, x = \frac{- 5 + 29}{2}, x = \frac{- 5 - 29}{2}

4 = (1 + \frac{r}{2})^{20}

9 x^{3} - 4 = 0

X^{3} - 64 X = 0

b^{4} = 339

x^{10} = 1000

What are the solutions to the equation x(x+5)=(x^2+5x)^2 ?
The solutions to the equation x(x+5)=(x^2+5x)^2 are x=0,x=-5,x=(-5+sqrt(29))/2 ,x=(-5-sqrt(29))/2
Find the zeros of x(x+5)=(x^2+5x)^2
The zeros of x(x+5)=(x^2+5x)^2 are x=0,x=-5,x=(-5+sqrt(29))/2 ,x=(-5-sqrt(29))/2