Precalculus Examples

$x^{2} + 6 x = - 19$

Step 1

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(3)}^{2}$

Step 2

Add the term to each side of the equation.

$x^{2} + 6 x + {(3)}^{2} = - 19 + {(3)}^{2}$

Step 3

Simplify the equation.

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Step 3.1

Simplify the left side.

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Step 3.1.1

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

$x^{2} + 6 x + 9 = - 19 + {(3)}^{2}$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $- 19 + {(3)}^{2}$ .

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Step 3.2.1.1

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

$x^{2} + 6 x + 9 = - 19 + 9$

Step 3.2.1.2

Add $- 19$ and $9$ .

$x^{2} + 6 x + 9 = - 10$

Step 4

Factor the perfect trinomial square into ${(x + 3)}^{2}$ .

${(x + 3)}^{2} = - 10$

Step 5

Solve the equation for $x$ .

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Step 5.1

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

$x + 3 = \pm \sqrt{- 10}$

Step 5.2

Simplify $\pm \sqrt{- 10}$ .

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Step 5.2.1

Rewrite $- 10$ as $- 1 (10)$ .

$x + 3 = \pm \sqrt{- 1 (10)}$

Step 5.2.2

Rewrite $\sqrt{- 1 (10)}$ as $\sqrt{- 1} \cdot \sqrt{10}$ .

$x + 3 = \pm \sqrt{- 1} \cdot \sqrt{10}$

Step 5.2.3

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

$x + 3 = \pm i \sqrt{10}$

Step 5.3

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

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Step 5.3.1

Subtract $3$ from both sides of the equation.

$x = \pm i \sqrt{10} - 3$

Step 5.3.2

Reorder $\pm i \sqrt{10}$ and $- 3$ .

$x = - 3 \pm i \sqrt{10}$