Precalculus Examples

$1211 = 900 {(1 + r)}^{2}$

Step 1

Rewrite the equation as $900 {(1 + r)}^{2} = 1211$ .

$900 {(1 + r)}^{2} = 1211$

Step 2

Divide each term in $900 {(1 + r)}^{2} = 1211$ by $900$ and simplify.

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Step 2.1

Divide each term in $900 {(1 + r)}^{2} = 1211$ by $900$ .

$\frac{900 {(1 + r)}^{2}}{900} = \frac{1211}{900}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $900$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{900 {(1 + r)}^{2}}{900} = \frac{1211}{900}$

Step 2.2.1.2

Divide ${(1 + r)}^{2}$ by $1$ .

${(1 + r)}^{2} = \frac{1211}{900}$

Step 3

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

$1 + r = \pm \sqrt{\frac{1211}{900}}$

Step 4

Simplify $\pm \sqrt{\frac{1211}{900}}$ .

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Step 4.1

Rewrite $\sqrt{\frac{1211}{900}}$ as $\frac{\sqrt{1211}}{\sqrt{900}}$ .

$1 + r = \pm \frac{\sqrt{1211}}{\sqrt{900}}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Rewrite $900$ as $30^{2}$ .

$1 + r = \pm \frac{\sqrt{1211}}{\sqrt{30^{2}}}$

Step 4.2.2

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

$1 + r = \pm \frac{\sqrt{1211}}{30}$

Step 5

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

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Step 5.1

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

$1 + r = \frac{\sqrt{1211}}{30}$

Step 5.2

Subtract $1$ from both sides of the equation.

$r = \frac{\sqrt{1211}}{30} - 1$

Step 5.3

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

$1 + r = - \frac{\sqrt{1211}}{30}$

Step 5.4

Subtract $1$ from both sides of the equation.

$r = - \frac{\sqrt{1211}}{30} - 1$

Step 5.5

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

$r = \frac{\sqrt{1211}}{30} - 1, - \frac{\sqrt{1211}}{30} - 1$

Step 6

The result can be shown in multiple forms.

Exact Form:

$r = \frac{\sqrt{1211}}{30} - 1, - \frac{\sqrt{1211}}{30} - 1$

Decimal Form:

$r = 0.15998084 \dots, - 2.15998084 \dots$