Solve by Completing the Square x^2+5x+6=0

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

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.

Simplify the left side.

Simplify each term.

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

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

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

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

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

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

Simplify the right side.

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

Simplify each term.

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 6$ by $4$ .

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

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 $x$ .

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

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

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

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

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

Any root of $1$ is $1$ .

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

Simplify the denominator.

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Subtract $5$ from $1$ .

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

Divide $- 4$ by $2$ .

$x = - 2$

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

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

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

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

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

Combine the numerators over the common denominator.

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

Subtract $5$ from $- 1$ .

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

Divide $- 6$ by $2$ .

$x = - 3$

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

$x = - 2, - 3$

Solve by Completing the Square x^2+5x+6=0

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