Solve by Completing the Square x^2-5x=-13

$x^{2} - 5 x = - 13$

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 $b$ .

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

Step 2

Add the term to each side of the equation.

$x^{2} - 5 x + {(- \frac{5}{2})}^{2} = - 13 + {(- \frac{5}{2})}^{2}$

Step 3

Simplify the equation.

Simplify the left side.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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

$x^{2} - 5 x + {(- 1)}^{2} {(\frac{5}{2})}^{2} = - 13 + {(- \frac{5}{2})}^{2}$

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

$x^{2} - 5 x + {(- 1)}^{2} \frac{5^{2}}{2^{2}} = - 13 + {(- \frac{5}{2})}^{2}$

Raise $- 1$ to the power of $2$ .

$x^{2} - 5 x + 1 \frac{5^{2}}{2^{2}} = - 13 + {(- \frac{5}{2})}^{2}$

Multiply $\frac{5^{2}}{2^{2}}$ by $1$ .

$x^{2} - 5 x + \frac{5^{2}}{2^{2}} = - 13 + {(- \frac{5}{2})}^{2}$

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

$x^{2} - 5 x + \frac{25}{2^{2}} = - 13 + {(- \frac{5}{2})}^{2}$

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

$x^{2} - 5 x + \frac{25}{4} = - 13 + {(- \frac{5}{2})}^{2}$

Simplify the right side.

Simplify $- 13 + {(- \frac{5}{2})}^{2}$ .

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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

$x^{2} - 5 x + \frac{25}{4} = - 13 + {(- 1)}^{2} {(\frac{5}{2})}^{2}$

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

$x^{2} - 5 x + \frac{25}{4} = - 13 + {(- 1)}^{2} \frac{5^{2}}{2^{2}}$

Raise $- 1$ to the power of $2$ .

$x^{2} - 5 x + \frac{25}{4} = - 13 + 1 \frac{5^{2}}{2^{2}}$

Multiply $\frac{5^{2}}{2^{2}}$ by $1$ .

$x^{2} - 5 x + \frac{25}{4} = - 13 + \frac{5^{2}}{2^{2}}$

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

$x^{2} - 5 x + \frac{25}{4} = - 13 + \frac{25}{2^{2}}$

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

$x^{2} - 5 x + \frac{25}{4} = - 13 + \frac{25}{4}$

To write $- 13$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x^{2} - 5 x + \frac{25}{4} = - 13 \cdot \frac{4}{4} + \frac{25}{4}$

Combine $- 13$ and $\frac{4}{4}$ .

$x^{2} - 5 x + \frac{25}{4} = \frac{- 13 \cdot 4}{4} + \frac{25}{4}$

Combine the numerators over the common denominator.

$x^{2} - 5 x + \frac{25}{4} = \frac{- 13 \cdot 4 + 25}{4}$

Simplify the numerator.

Multiply $- 13$ by $4$ .

$x^{2} - 5 x + \frac{25}{4} = \frac{- 52 + 25}{4}$

Add $- 52$ and $25$ .

$x^{2} - 5 x + \frac{25}{4} = \frac{- 27}{4}$

Move the negative in front of the fraction.

$x^{2} - 5 x + \frac{25}{4} = - \frac{27}{4}$

Step 4

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

${(x - \frac{5}{2})}^{2} = - \frac{27}{4}$

Step 5

Solve the equation for $x$ .

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

$x - \frac{5}{2} = \pm \sqrt{- \frac{27}{4}}$

Simplify $\pm \sqrt{- \frac{27}{4}}$ .

Rewrite $- \frac{27}{4}$ as ${(\frac{3 i}{2})}^{2} \cdot 3$ .

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

$x - \frac{5}{2} = \pm \sqrt{i^{2} \frac{27}{4}}$

Factor the perfect power $3^{2}$ out of $27$ .

$x - \frac{5}{2} = \pm \sqrt{i^{2} \frac{3^{2} \cdot 3}{4}}$

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

$x - \frac{5}{2} = \pm \sqrt{i^{2} \frac{3^{2} \cdot 3}{2^{2} \cdot 1}}$

Rearrange the fraction $\frac{3^{2} \cdot 3}{2^{2} \cdot 1}$ .

$x - \frac{5}{2} = \pm \sqrt{i^{2} ({(\frac{3}{2})}^{2} \cdot 3)}$

Rewrite $i^{2} {(\frac{3}{2})}^{2}$ as ${(\frac{3 i}{2})}^{2}$ .

$x - \frac{5}{2} = \pm \sqrt{{(\frac{3 i}{2})}^{2} \cdot 3}$

Pull terms out from under the radical.

$x - \frac{5}{2} = \pm \frac{3 i}{2} \sqrt{3}$

Combine $\frac{3 i}{2}$ and $\sqrt{3}$ .

$x - \frac{5}{2} = \pm \frac{3 i \sqrt{3}}{2}$

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

Add $\frac{5}{2}$ to both sides of the equation.

$x = \pm \frac{3 i \sqrt{3}}{2} + \frac{5}{2}$

Reorder $\pm \frac{3 i \sqrt{3}}{2}$ and $\frac{5}{2}$ .

$x = \frac{5}{2} \pm \frac{3 i \sqrt{3}}{2}$

Solve by Completing the Square x^2-5x=-13

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