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

The solutions to the equation x(x-1)(x+2)= 1/4 x are x=0,x=(-1+sqrt(10))/2 ,x=-(1+sqrt(10))/2

Find the zeros of x(x-1)(x+2)= 1/4 x

The zeros of x(x-1)(x+2)= 1/4 x are x=0,x=(-1+sqrt(10))/2 ,x=-(1+sqrt(10))/2

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

x = 0, x = \frac{- 1 + 10}{2}, x = - \frac{1 + 10}{2}

Solution steps

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

Expand

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

x^{3} + x^{2} - 2 x = \frac{1}{4} x

Move

\frac{1}{4} x

to the left side

x^{3} + x^{2} - \frac{9}{4} x = 0

Factor

x^{3} + x^{2} - \frac{9}{4} x : x (x^{2} + x - \frac{9}{4})

x (x^{2} + x - \frac{9}{4}) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x^{2} + x - \frac{9}{4} = 0

Solve

x^{2} + x - \frac{9}{4} = 0 : x = \frac{- 1 + 10}{2}, x = - \frac{1 + 10}{2}

The solutions are

x = 0, x = \frac{- 1 + 10}{2}, x = - \frac{1 + 10}{2}

x^{15} = \frac{- 6 x - 9}{3}

x^{6} + 6 x^{3} - 16 = 0

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

y, x^{2} = y^{3}

2 x^{3} + 32 = 8 x^{2} + 8 x

What are the solutions to the equation x(x-1)(x+2)= 1/4 x ?
The solutions to the equation x(x-1)(x+2)= 1/4 x are x=0,x=(-1+sqrt(10))/2 ,x=-(1+sqrt(10))/2
Find the zeros of x(x-1)(x+2)= 1/4 x
The zeros of x(x-1)(x+2)= 1/4 x are x=0,x=(-1+sqrt(10))/2 ,x=-(1+sqrt(10))/2