Solve (x^2 3x 1) (x^2 3x - 3) ≥5

Evaluate

(x^{2} \times 3 x \times 1) (x^{2} \times 3 x - 3) \geq 5

Remove the parentheses

x^{2} \times 3 x \times 1 \times (x^{2} \times 3 x - 3) \geq 5

Simplify

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3 x^{3} (3 x^{3} - 3) \geq 5

Move the expression to the left side

3 x^{3} (3 x^{3} - 3) - 5 \geq 0

Expand the expression

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9 x^{6} - 9 x^{3} - 5 \geq 0

Rewrite the expression

9 x^{6} - 9 x^{3} - 5 = 0

Solve the equation using substitution t = x^{3}

9 t^{2} - 9 t - 5 = 0

Add or subtract both sides

9 t^{2} - 9 t = 5

Divide both sides

\frac{9 t ^{2} - 9 t}{9} = \frac{5}{9}

Evaluate

t^{2} - t = \frac{5}{9}

Add the same value to both sides

t^{2} - t + \frac{1}{4} = \frac{5}{9} + \frac{1}{4}

Simplify the expression

(t - \frac{1}{2})^{2} = \frac{29}{36}

Take the root of both sides of the equation and remember to use both positive and negative roots

t - \frac{1}{2} = \pm \frac{29}{36}

Simplify the expression

t - \frac{1}{2} = \pm \frac{29}{6}

Separate the equation into 2 possible cases

t - \frac{1}{2} = \frac{29}{6} t - \frac{1}{2} = - \frac{29}{6}

Solve the equation

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t = \frac{29 + 3}{6} t - \frac{1}{2} = - \frac{29}{6}

Solve the equation

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t = \frac{29 + 3}{6} t = \frac{- 29 + 3}{6}

Substitute back

x^{3} = \frac{29 + 3}{6} x^{3} = \frac{- 29 + 3}{6}

Solve the equation for x

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x = \frac{3 36 29 + 108}{6} x^{3} = \frac{- 29 + 3}{6}

Solve the equation for x

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x = \frac{3 36 29 + 108}{6} x = - \frac{3 36 29 - 108}{6}

Determine the test intervals using the critical values

x < - \frac{3 36 29 - 108}{6} - \frac{3 36 29 - 108}{6} < x < \frac{3 36 29 + 108}{6} x > \frac{3 36 29 + 108}{6}

Choose a value form each interval

x_{1} = - 2 x_{2} = 0 x_{3} = 2

To determine if x < - \frac{3 36 29 - 108}{6} is the solution to the inequality,test if the chosen value x = - 2 satisfies the initial inequality

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x < - \frac{3 36 29 - 108}{6} is the solution x_{2} = 0 x_{3} = 2

To determine if - \frac{3 36 29 - 108}{6} < x < \frac{3 36 29 + 108}{6} is the solution to the inequality,test if the chosen value x = 0 satisfies the initial inequality

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x < - \frac{3 36 29 - 108}{6} is the solution - \frac{3 36 29 - 108}{6} < x < \frac{3 36 29 + 108}{6} is not a solution x_{3} = 2

To determine if x > \frac{3 36 29 + 108}{6} is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

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x < - \frac{3 36 29 - 108}{6} is the solution - \frac{3 36 29 - 108}{6} < x < \frac{3 36 29 + 108}{6} is not a solution x > \frac{3 36 29 + 108}{6} is the solution

The original inequality is a nonstrict inequality,so include the critical value in the solution

x \leq - \frac{3 36 29 - 108}{6} is the solution x \geq \frac{3 36 29 + 108}{6} is the solution

Solution

x \in - \infty, - \frac{3 36 29 - 108}{6} \cup \frac{3 36 29 + 108}{6}, + \infty

Solve the inequality