Solve by Completing the Square x^2+5x+8=0

$x^{2} + 5 x + 8 = 0$

Step 1

Subtract $8$ from both sides of the equation.

$x^{2} + 5 x = - 8$

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.

$x^{2} + 5 x + {(\frac{5}{2})}^{2} = - 8 + {(\frac{5}{2})}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Simplify each term.

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

$x^{2} + 5 x + \frac{5^{2}}{2^{2}} = - 8 + {(\frac{5}{2})}^{2}$

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

$x^{2} + 5 x + \frac{25}{2^{2}} = - 8 + {(\frac{5}{2})}^{2}$

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

$x^{2} + 5 x + \frac{25}{4} = - 8 + {(\frac{5}{2})}^{2}$

Simplify the right side.

Simplify $- 8 + {(\frac{5}{2})}^{2}$ .

Simplify each term.

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

$x^{2} + 5 x + \frac{25}{4} = - 8 + \frac{5^{2}}{2^{2}}$

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

$x^{2} + 5 x + \frac{25}{4} = - 8 + \frac{25}{2^{2}}$

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

$x^{2} + 5 x + \frac{25}{4} = - 8 + \frac{25}{4}$

To write $- 8$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x^{2} + 5 x + \frac{25}{4} = - 8 \cdot \frac{4}{4} + \frac{25}{4}$

Combine $- 8$ and $\frac{4}{4}$ .

$x^{2} + 5 x + \frac{25}{4} = \frac{- 8 \cdot 4}{4} + \frac{25}{4}$

Combine the numerators over the common denominator.

$x^{2} + 5 x + \frac{25}{4} = \frac{- 8 \cdot 4 + 25}{4}$

Simplify the numerator.

Multiply $- 8$ by $4$ .

$x^{2} + 5 x + \frac{25}{4} = \frac{- 32 + 25}{4}$

Add $- 32$ and $25$ .

$x^{2} + 5 x + \frac{25}{4} = \frac{- 7}{4}$

Move the negative in front of the fraction.

$x^{2} + 5 x + \frac{25}{4} = - \frac{7}{4}$

Step 5

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

${(x + \frac{5}{2})}^{2} = - \frac{7}{4}$

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.

$x + \frac{5}{2} = \pm \sqrt{- \frac{7}{4}}$

Simplify $\pm \sqrt{- \frac{7}{4}}$ .

Rewrite $- \frac{7}{4}$ as ${(\frac{i}{2})}^{2} \cdot 7$ .

Rewrite $- 1$ as $i^{2}$ .

$x + \frac{5}{2} = \pm \sqrt{i^{2} \frac{7}{4}}$

Factor the perfect power $1^{2}$ out of $7$ .

$x + \frac{5}{2} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 7}{4}}$

Factor the perfect power $2^{2}$ out of $4$ .

$x + \frac{5}{2} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 7}{2^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} \cdot 7}{2^{2} \cdot 1}$ .

$x + \frac{5}{2} = \pm \sqrt{i^{2} ({(\frac{1}{2})}^{2} \cdot 7)}$

Rewrite $i^{2} {(\frac{1}{2})}^{2}$ as ${(\frac{i}{2})}^{2}$ .

$x + \frac{5}{2} = \pm \sqrt{{(\frac{i}{2})}^{2} \cdot 7}$

Pull terms out from under the radical.

$x + \frac{5}{2} = \pm \frac{i}{2} \sqrt{7}$

Combine $\frac{i}{2}$ and $\sqrt{7}$ .

$x + \frac{5}{2} = \pm \frac{i \sqrt{7}}{2}$

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

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

$x = \pm \frac{i \sqrt{7}}{2} - \frac{5}{2}$

Reorder $\pm \frac{i \sqrt{7}}{2}$ and $- \frac{5}{2}$ .

$x = - \frac{5}{2} \pm \frac{i \sqrt{7}}{2}$

Solve by Completing the Square x^2+5x+8=0

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