Precalculus Examples

$x^{3} - x^{2} \geq - 9 x + 9$

Step 1

Add $9 x$ to both sides of the inequality.

$x^{3} - x^{2} + 9 x \geq 9$

Step 2

Convert the inequality to an equation.

$x^{3} - x^{2} + 9 x = 9$

Step 3

Subtract $9$ from both sides of the equation.

$x^{3} - x^{2} + 9 x - 9 = 0$

Step 4

Factor the left side of the equation.

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Step 4.1

Factor out the greatest common factor from each group.

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Step 4.1.1

Group the first two terms and the last two terms.

$(x^{3} - x^{2}) + 9 x - 9 = 0$

Step 4.1.2

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x - 1) + 9 (x - 1) = 0$

Step 4.2

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$(x - 1) (x^{2} + 9) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$x^{2} + 9 = 0$

Step 6

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 6.2

Add $1$ to both sides of the equation.

$x = 1$

Step 7

Set $x^{2} + 9$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x^{2} + 9$ equal to $0$ .

$x^{2} + 9 = 0$

Step 7.2

Solve $x^{2} + 9 = 0$ for $x$ .

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Step 7.2.1

Subtract $9$ from both sides of the equation.

$x^{2} = - 9$

Step 7.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 9}$

Step 7.2.3

Simplify $\pm \sqrt{- 9}$ .

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Step 7.2.3.1

Rewrite $- 9$ as $- 1 (9)$ .

$x = \pm \sqrt{- 1 (9)}$

Step 7.2.3.2

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{9}$

Step 7.2.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{9}$

Step 7.2.3.4

Rewrite $9$ as $3^{2}$ .

$x = \pm i \cdot \sqrt{3^{2}}$

Step 7.2.3.5

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 3$

Step 7.2.3.6

Move $3$ to the left of $i$ .

$x = \pm 3 i$

Step 7.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 7.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 3 i$

Step 7.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3 i$

Step 7.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3 i, - 3 i$

Step 8

The final solution is all the values that make $(x - 1) (x^{2} + 9) = 0$ true.

$x = 1, 3 i, - 3 i$

Step 9

The solution consists of all of the true intervals.

$x \geq 1$

Step 10

Convert the inequality to interval notation.

$[1, \infty)$

Step 11