Solve by Completing the Square 2x^2-5x+67=0

$2 x^{2} - 5 x + 67 = 0$

Step 1

Subtract $67$ from both sides of the equation.

$2 x^{2} - 5 x = - 67$

Step 2

Divide each term in $2 x^{2} - 5 x = - 67$ by $2$ and simplify.

Divide each term in $2 x^{2} - 5 x = - 67$ by $2$ .

$\frac{2 x^{2}}{2} + \frac{- 5 x}{2} = \frac{- 67}{2}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} + \frac{- 5 x}{2} = \frac{- 67}{2}$

Divide $x^{2}$ by $1$ .

$x^{2} + \frac{- 5 x}{2} = \frac{- 67}{2}$

Move the negative in front of the fraction.

$x^{2} - \frac{5 x}{2} = \frac{- 67}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x^{2} - \frac{5 x}{2} = - \frac{67}{2}$

Step 3

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

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

Step 4

Add the term to each side of the equation.

$x^{2} - \frac{5 x}{2} + {(- \frac{5}{4})}^{2} = - \frac{67}{2} + {(- \frac{5}{4})}^{2}$

Step 5

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 $- \frac{5}{4}$ .

$x^{2} - \frac{5 x}{2} + {(- 1)}^{2} {(\frac{5}{4})}^{2} = - \frac{67}{2} + {(- \frac{5}{4})}^{2}$

Apply the product rule to $\frac{5}{4}$ .

$x^{2} - \frac{5 x}{2} + {(- 1)}^{2} \frac{5^{2}}{4^{2}} = - \frac{67}{2} + {(- \frac{5}{4})}^{2}$

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

$x^{2} - \frac{5 x}{2} + 1 \frac{5^{2}}{4^{2}} = - \frac{67}{2} + {(- \frac{5}{4})}^{2}$

Multiply $\frac{5^{2}}{4^{2}}$ by $1$ .

$x^{2} - \frac{5 x}{2} + \frac{5^{2}}{4^{2}} = - \frac{67}{2} + {(- \frac{5}{4})}^{2}$

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

$x^{2} - \frac{5 x}{2} + \frac{25}{4^{2}} = - \frac{67}{2} + {(- \frac{5}{4})}^{2}$

Raise $4$ to the power of $2$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{67}{2} + {(- \frac{5}{4})}^{2}$

Simplify the right side.

Simplify $- \frac{67}{2} + {(- \frac{5}{4})}^{2}$ .

Simplify each term.

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

Apply the product rule to $- \frac{5}{4}$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{67}{2} + {(- 1)}^{2} {(\frac{5}{4})}^{2}$

Apply the product rule to $\frac{5}{4}$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{67}{2} + {(- 1)}^{2} \frac{5^{2}}{4^{2}}$

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

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{67}{2} + 1 \frac{5^{2}}{4^{2}}$

Multiply $\frac{5^{2}}{4^{2}}$ by $1$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{67}{2} + \frac{5^{2}}{4^{2}}$

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

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{67}{2} + \frac{25}{4^{2}}$

Raise $4$ to the power of $2$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{67}{2} + \frac{25}{16}$

To write $- \frac{67}{2}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{67}{2} \cdot \frac{8}{8} + \frac{25}{16}$

Write each expression with a common denominator of $16$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{67}{2}$ by $\frac{8}{8}$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{67 \cdot 8}{2 \cdot 8} + \frac{25}{16}$

Multiply $2$ by $8$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{67 \cdot 8}{16} + \frac{25}{16}$

Combine the numerators over the common denominator.

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = \frac{- 67 \cdot 8 + 25}{16}$

Simplify the numerator.

Multiply $- 67$ by $8$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = \frac{- 536 + 25}{16}$

Add $- 536$ and $25$ .

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = \frac{- 511}{16}$

Move the negative in front of the fraction.

$x^{2} - \frac{5 x}{2} + \frac{25}{16} = - \frac{511}{16}$

Step 6

Factor the perfect trinomial square into ${(x - \frac{5}{4})}^{2}$ .

${(x - \frac{5}{4})}^{2} = - \frac{511}{16}$

Step 7

Solve the equation for $x$ .

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

$x - \frac{5}{4} = \pm \sqrt{- \frac{511}{16}}$

Simplify $\pm \sqrt{- \frac{511}{16}}$ .

Rewrite $- \frac{511}{16}$ as ${(\frac{i}{4})}^{2} \cdot 511$ .

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

$x - \frac{5}{4} = \pm \sqrt{i^{2} \frac{511}{16}}$

Factor the perfect power $1^{2}$ out of $511$ .

$x - \frac{5}{4} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 511}{16}}$

Factor the perfect power $4^{2}$ out of $16$ .

$x - \frac{5}{4} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 511}{4^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} \cdot 511}{4^{2} \cdot 1}$ .

$x - \frac{5}{4} = \pm \sqrt{i^{2} ({(\frac{1}{4})}^{2} \cdot 511)}$

Rewrite $i^{2} {(\frac{1}{4})}^{2}$ as ${(\frac{i}{4})}^{2}$ .

$x - \frac{5}{4} = \pm \sqrt{{(\frac{i}{4})}^{2} \cdot 511}$

Pull terms out from under the radical.

$x - \frac{5}{4} = \pm \frac{i}{4} \sqrt{511}$

Combine $\frac{i}{4}$ and $\sqrt{511}$ .

$x - \frac{5}{4} = \pm \frac{i \sqrt{511}}{4}$

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

Add $\frac{5}{4}$ to both sides of the equation.

$x = \pm \frac{i \sqrt{511}}{4} + \frac{5}{4}$

Reorder $\pm \frac{i \sqrt{511}}{4}$ and $\frac{5}{4}$ .

$x = \frac{5}{4} \pm \frac{i \sqrt{511}}{4}$

Solve by Completing the Square 2x^2-5x+67=0

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