Algebra Examples

$x^{2} + 5 x + 6 = 0$

Step 1

Subtract $6$ from both sides of the equation.

$x^{2} + 5 x = - 6$

Step 2

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

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

Step 3

Add the term to each side of the equation.

$x^{2} + 5 x + {(\frac{5}{2})}^{2} = - 6 + {(\frac{5}{2})}^{2}$

Step 4

Simplify the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Simplify each term.

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Step 4.1.1.1

Apply the product rule to $\frac{5}{2}$ .

$x^{2} + 5 x + \frac{5^{2}}{2^{2}} = - 6 + {(\frac{5}{2})}^{2}$

Step 4.1.1.2

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

$x^{2} + 5 x + \frac{25}{2^{2}} = - 6 + {(\frac{5}{2})}^{2}$

Step 4.1.1.3

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

$x^{2} + 5 x + \frac{25}{4} = - 6 + {(\frac{5}{2})}^{2}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $- 6 + {(\frac{5}{2})}^{2}$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Apply the product rule to $\frac{5}{2}$ .

$x^{2} + 5 x + \frac{25}{4} = - 6 + \frac{5^{2}}{2^{2}}$

Step 4.2.1.1.2

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

$x^{2} + 5 x + \frac{25}{4} = - 6 + \frac{25}{2^{2}}$

Step 4.2.1.1.3

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

$x^{2} + 5 x + \frac{25}{4} = - 6 + \frac{25}{4}$

Step 4.2.1.2

To write $- 6$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x^{2} + 5 x + \frac{25}{4} = - 6 \cdot \frac{4}{4} + \frac{25}{4}$

Step 4.2.1.3

Combine $- 6$ and $\frac{4}{4}$ .

$x^{2} + 5 x + \frac{25}{4} = \frac{- 6 \cdot 4}{4} + \frac{25}{4}$

Step 4.2.1.4

Combine the numerators over the common denominator.

$x^{2} + 5 x + \frac{25}{4} = \frac{- 6 \cdot 4 + 25}{4}$

Step 4.2.1.5

Simplify the numerator.

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Step 4.2.1.5.1

Multiply $- 6$ by $4$ .

$x^{2} + 5 x + \frac{25}{4} = \frac{- 24 + 25}{4}$

Step 4.2.1.5.2

Add $- 24$ and $25$ .

$x^{2} + 5 x + \frac{25}{4} = \frac{1}{4}$

Step 5

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

${(x + \frac{5}{2})}^{2} = \frac{1}{4}$

Step 6

Solve the equation for $x$ .

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Step 6.1

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

$x + \frac{5}{2} = \pm \sqrt{\frac{1}{4}}$

Step 6.2

Simplify $\pm \sqrt{\frac{1}{4}}$ .

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Step 6.2.1

Rewrite $\sqrt{\frac{1}{4}}$ as $\frac{\sqrt{1}}{\sqrt{4}}$ .

$x + \frac{5}{2} = \pm \frac{\sqrt{1}}{\sqrt{4}}$

Step 6.2.2

Any root of $1$ is $1$ .

$x + \frac{5}{2} = \pm \frac{1}{\sqrt{4}}$

Step 6.2.3

Simplify the denominator.

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Step 6.2.3.1

Rewrite $4$ as $2^{2}$ .

$x + \frac{5}{2} = \pm \frac{1}{\sqrt{2^{2}}}$

Step 6.2.3.2

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

$x + \frac{5}{2} = \pm \frac{1}{2}$

Step 6.3

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

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Step 6.3.1

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

$x + \frac{5}{2} = \frac{1}{2}$

Step 6.3.2

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

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Step 6.3.2.1

Subtract $\frac{5}{2}$ from both sides of the equation.

$x = \frac{1}{2} - \frac{5}{2}$

Step 6.3.2.2

Combine the numerators over the common denominator.

$x = \frac{1 - 5}{2}$

Step 6.3.2.3

Subtract $5$ from $1$ .

$x = \frac{- 4}{2}$

Step 6.3.2.4

Divide $- 4$ by $2$ .

$x = - 2$

Step 6.3.3

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

$x + \frac{5}{2} = - \frac{1}{2}$

Step 6.3.4

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

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Step 6.3.4.1

Subtract $\frac{5}{2}$ from both sides of the equation.

$x = - \frac{1}{2} - \frac{5}{2}$

Step 6.3.4.2

Combine the numerators over the common denominator.

$x = \frac{- 1 - 5}{2}$

Step 6.3.4.3

Subtract $5$ from $- 1$ .

$x = \frac{- 6}{2}$

Step 6.3.4.4

Divide $- 6$ by $2$ .

$x = - 3$

Step 6.3.5

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

$x = - 2, - 3$