Precalculus Examples

$x^{2} + 10 x = - 24$

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 $b$ .

${(\frac{b}{2})}^{2} = {(5)}^{2}$

Step 2

Add the term to each side of the equation.

$x^{2} + 10 x + {(5)}^{2} = - 24 + {(5)}^{2}$

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 $5$ to the power of $2$ .

$x^{2} + 10 x + 25 = - 24 + {(5)}^{2}$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $- 24 + {(5)}^{2}$ .

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Step 3.2.1.1

Raise $5$ to the power of $2$ .

$x^{2} + 10 x + 25 = - 24 + 25$

Step 3.2.1.2

Add $- 24$ and $25$ .

$x^{2} + 10 x + 25 = 1$

Step 4

Factor the perfect trinomial square into ${(x + 5)}^{2}$ .

${(x + 5)}^{2} = 1$

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.

$x + 5 = \pm \sqrt{1}$

Step 5.2

Any root of $1$ is $1$ .

$x + 5 = \pm 1$

Step 5.3

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

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Step 5.3.1

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

$x + 5 = 1$

Step 5.3.2

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

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Step 5.3.2.1

Subtract $5$ from both sides of the equation.

$x = 1 - 5$

Step 5.3.2.2

Subtract $5$ from $1$ .

$x = - 4$

Step 5.3.3

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

$x + 5 = - 1$

Step 5.3.4

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

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Step 5.3.4.1

Subtract $5$ from both sides of the equation.

$x = - 1 - 5$

Step 5.3.4.2

Subtract $5$ from $- 1$ .

$x = - 6$

Step 5.3.5

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

$x = - 4, - 6$