Precalculus Examples

${(x + 3)}^{2} - 169 = 0$

Step 1

Add $169$ to both sides of the equation.

${(x + 3)}^{2} = 169$

Step 2

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

$x + 3 = \pm \sqrt{169}$

Step 3

Simplify $\pm \sqrt{169}$ .

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Step 3.1

Rewrite $169$ as $13^{2}$ .

$x + 3 = \pm \sqrt{13^{2}}$

Step 3.2

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

$x + 3 = \pm 13$

Step 4

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

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Step 4.1

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

$x + 3 = 13$

Step 4.2

Move all terms not containing $x$ to the right side of the equation.

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Step 4.2.1

Subtract $3$ from both sides of the equation.

$x = 13 - 3$

Step 4.2.2

Subtract $3$ from $13$ .

$x = 10$

Step 4.3

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

$x + 3 = - 13$

Step 4.4

Move all terms not containing $x$ to the right side of the equation.

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Step 4.4.1

Subtract $3$ from both sides of the equation.

$x = - 13 - 3$

Step 4.4.2

Subtract $3$ from $- 13$ .

$x = - 16$

Step 4.5

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

$x = 10, - 16$