Solve |x^(2) 3x 9|-x^(2)-x

Question

Simplify the expression

27 x^{3} - x^{2} - x

Show Solution

x_{1} = 0, x_{2} = \frac{1 + 109}{54}

Alternative Form

x_{1} = 0, x_{2} \approx 0.211858

Evaluate

x^{2} \times 3 x \times 9 - x^{2} - x

To find the roots of the expression,set the expression equal to 0

x^{2} \times 3 x \times 9 - x^{2} - x = 0

Multiply

More Steps

27 x^{3} - x^{2} - x = 0

Calculate the absolute value

27 x^{3} - x^{2} - x = 0

Separate the equation into 2 possible cases

27 x^{3} - x^{2} - x = 0, x^{3} \geq 0 27 (- x^{3}) - x^{2} - x = 0, x^{3} < 0

Solve the equation

More Steps

x = 0 x = \frac{1 + 109}{54} x = \frac{1 - 109}{54}, x^{3} \geq 0 27 (- x^{3}) - x^{2} - x = 0, x^{3} < 0

The only way a base raised to an odd power can be greater than or equal to 0 is if the base is greater than or equal to 0

x = 0 x = \frac{1 + 109}{54} x = \frac{1 - 109}{54}, x \geq 0 27 (- x^{3}) - x^{2} - x = 0, x^{3} < 0

Solve the equation

More Steps

x = 0 x = \frac{1 + 109}{54} x = \frac{1 - 109}{54}, x \geq 0 x = 0 x = - \frac{1}{54} + \frac{107}{54} i x = - \frac{1}{54} - \frac{107}{54} i, x^{3} < 0

The only way a base raised to an odd power can be less than 0 is if the base is less than 0

x = 0 x = \frac{1 + 109}{54} x = \frac{1 - 109}{54}, x \geq 0 x = 0 x = - \frac{1}{54} + \frac{107}{54} i x = - \frac{1}{54} - \frac{107}{54} i, x < 0

Find the intersection

x = 0 x = \frac{1 + 109}{54} x = 0 x = - \frac{1}{54} + \frac{107}{54} i x = - \frac{1}{54} - \frac{107}{54} i, x < 0

Find the intersection

x = 0 x = \frac{1 + 109}{54} x \in \emptyset

Find the union

x = 0 x = \frac{1 + 109}{54}

Solution

x_{1} = 0, x_{2} = \frac{1 + 109}{54}

Alternative Form

x_{1} = 0, x_{2} \approx 0.211858

Show Solution