Finite Math Examples

$\frac{a^{2} r^{6}}{a^{2} r^{3}} = \frac{729}{- 27}$

Step 1

Factor each term.

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Step 1.1

Reduce the expression $\frac{a^{2} r^{6}}{a^{2} r^{3}}$ by cancelling the common factors.

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Step 1.1.1

Cancel the common factor.

$\frac{a^{2} r^{6}}{a^{2} r^{3}} = \frac{729}{- 27}$

Step 1.1.2

Rewrite the expression.

$\frac{r^{6}}{r^{3}} = \frac{729}{- 27}$

Step 1.2

Cancel the common factor of $r^{6}$ and $r^{3}$ .

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Step 1.2.1

Factor $r^{3}$ out of $r^{6}$ .

$\frac{r^{3} r^{3}}{r^{3}} = \frac{729}{- 27}$

Step 1.2.2

Cancel the common factors.

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Step 1.2.2.1

Multiply by $1$ .

$\frac{r^{3} r^{3}}{r^{3} \cdot 1} = \frac{729}{- 27}$

Step 1.2.2.2

Cancel the common factor.

$\frac{r^{3} r^{3}}{r^{3} \cdot 1} = \frac{729}{- 27}$

Step 1.2.2.3

Rewrite the expression.

$\frac{r^{3}}{1} = \frac{729}{- 27}$

Step 1.2.2.4

Divide $r^{3}$ by $1$ .

$r^{3} = \frac{729}{- 27}$

Step 1.3

Divide $729$ by $- 27$ .

$r^{3} = - 27$

Step 2

Solve the equation.

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Step 2.1

Add $27$ to both sides of the equation.

$r^{3} + 27 = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Rewrite $27$ as $3^{3}$ .

$r^{3} + 3^{3} = 0$

Step 2.2.2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = r$ and $b = 3$ .

$(r + 3) (r^{2} - r \cdot 3 + 3^{2}) = 0$

Step 2.2.3

Simplify.

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Step 2.2.3.1

Multiply $3$ by $- 1$ .

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

Step 2.2.3.2

Raise $3$ to the power of $2$ .

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

Step 2.3

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

$r + 3 = 0$

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

Step 2.4

Set $r + 3$ equal to $0$ and solve for $r$ .

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Step 2.4.1

Set $r + 3$ equal to $0$ .

$r + 3 = 0$

Step 2.4.2

Subtract $3$ from both sides of the equation.

$r = - 3$

Step 2.5

Set $r^{2} - 3 r + 9$ equal to $0$ and solve for $r$ .

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Step 2.5.1

Set $r^{2} - 3 r + 9$ equal to $0$ .

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

Step 2.5.2

Solve $r^{2} - 3 r + 9 = 0$ for $r$ .

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Step 2.5.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.5.2.2

Substitute the values $a = 1$ , $b = - 3$ , and $c = 9$ into the quadratic formula and solve for $r$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Step 2.5.2.3

Simplify.

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Step 2.5.2.3.1

Simplify the numerator.

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Step 2.5.2.3.1.1

Raise $- 3$ to the power of $2$ .

$r = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Step 2.5.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 9$ .

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Step 2.5.2.3.1.2.1

Multiply $- 4$ by $1$ .

$r = \frac{3 \pm \sqrt{9 - 4 \cdot 9}}{2 \cdot 1}$

Step 2.5.2.3.1.2.2

Multiply $- 4$ by $9$ .

$r = \frac{3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Step 2.5.2.3.1.3

Subtract $36$ from $9$ .

$r = \frac{3 \pm \sqrt{- 27}}{2 \cdot 1}$

Step 2.5.2.3.1.4

Rewrite $- 27$ as $- 1 (27)$ .

$r = \frac{3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Step 2.5.2.3.1.5

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$r = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

Step 2.5.2.3.1.6

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

$r = \frac{3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Step 2.5.2.3.1.7

Rewrite $27$ as $3^{2} \cdot 3$ .

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Step 2.5.2.3.1.7.1

Factor $9$ out of $27$ .

$r = \frac{3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

Step 2.5.2.3.1.7.2

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

$r = \frac{3 \pm i \cdot \sqrt{3^{2} \cdot 3}}{2 \cdot 1}$

Step 2.5.2.3.1.8

Pull terms out from under the radical.

$r = \frac{3 \pm i \cdot (3 \sqrt{3})}{2 \cdot 1}$

Step 2.5.2.3.1.9

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

$r = \frac{3 \pm 3 i \sqrt{3}}{2 \cdot 1}$

Step 2.5.2.3.2

Multiply $2$ by $1$ .

$r = \frac{3 \pm 3 i \sqrt{3}}{2}$

Step 2.5.2.4

The final answer is the combination of both solutions.

$r = \frac{3 + 3 i \sqrt{3}}{2}, \frac{3 - 3 i \sqrt{3}}{2}$

Step 2.6

The final solution is all the values that make $(r + 3) (r^{2} - 3 r + 9) = 0$ true.

$r = - 3, \frac{3 + 3 i \sqrt{3}}{2}, \frac{3 - 3 i \sqrt{3}}{2}$

Step 3

The variable $a$ got canceled.

All real numbers

Step 4

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$