Solve by Completing the Square x^2-3x+4=0

$x^{2} - 3 x + 4 = 0$

Step 1

Subtract $4$ from both sides of the equation.

$x^{2} - 3 x = - 4$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{3}{2})}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} - 3 x + {(- \frac{3}{2})}^{2} = - 4 + {(- \frac{3}{2})}^{2}$

Step 4

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 $- \frac{3}{2}$ .

$x^{2} - 3 x + {(- 1)}^{2} {(\frac{3}{2})}^{2} = - 4 + {(- \frac{3}{2})}^{2}$

Apply the product rule to $\frac{3}{2}$ .

$x^{2} - 3 x + {(- 1)}^{2} \frac{3^{2}}{2^{2}} = - 4 + {(- \frac{3}{2})}^{2}$

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

$x^{2} - 3 x + 1 \frac{3^{2}}{2^{2}} = - 4 + {(- \frac{3}{2})}^{2}$

Multiply $\frac{3^{2}}{2^{2}}$ by $1$ .

$x^{2} - 3 x + \frac{3^{2}}{2^{2}} = - 4 + {(- \frac{3}{2})}^{2}$

Raise $3$ to the power of $2$ .

$x^{2} - 3 x + \frac{9}{2^{2}} = - 4 + {(- \frac{3}{2})}^{2}$

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

$x^{2} - 3 x + \frac{9}{4} = - 4 + {(- \frac{3}{2})}^{2}$

Simplify the right side.

Simplify $- 4 + {(- \frac{3}{2})}^{2}$ .

Simplify each term.

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

Apply the product rule to $- \frac{3}{2}$ .

$x^{2} - 3 x + \frac{9}{4} = - 4 + {(- 1)}^{2} {(\frac{3}{2})}^{2}$

Apply the product rule to $\frac{3}{2}$ .

$x^{2} - 3 x + \frac{9}{4} = - 4 + {(- 1)}^{2} \frac{3^{2}}{2^{2}}$

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

$x^{2} - 3 x + \frac{9}{4} = - 4 + 1 \frac{3^{2}}{2^{2}}$

Multiply $\frac{3^{2}}{2^{2}}$ by $1$ .

$x^{2} - 3 x + \frac{9}{4} = - 4 + \frac{3^{2}}{2^{2}}$

Raise $3$ to the power of $2$ .

$x^{2} - 3 x + \frac{9}{4} = - 4 + \frac{9}{2^{2}}$

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

$x^{2} - 3 x + \frac{9}{4} = - 4 + \frac{9}{4}$

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

$x^{2} - 3 x + \frac{9}{4} = - 4 \cdot \frac{4}{4} + \frac{9}{4}$

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

$x^{2} - 3 x + \frac{9}{4} = \frac{- 4 \cdot 4}{4} + \frac{9}{4}$

Combine the numerators over the common denominator.

$x^{2} - 3 x + \frac{9}{4} = \frac{- 4 \cdot 4 + 9}{4}$

Simplify the numerator.

Multiply $- 4$ by $4$ .

$x^{2} - 3 x + \frac{9}{4} = \frac{- 16 + 9}{4}$

Add $- 16$ and $9$ .

$x^{2} - 3 x + \frac{9}{4} = \frac{- 7}{4}$

Move the negative in front of the fraction.

$x^{2} - 3 x + \frac{9}{4} = - \frac{7}{4}$

Step 5

Factor the perfect trinomial square into ${(x - \frac{3}{2})}^{2}$ .

${(x - \frac{3}{2})}^{2} = - \frac{7}{4}$

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.

$x - \frac{3}{2} = \pm \sqrt{- \frac{7}{4}}$

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

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

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

$x - \frac{3}{2} = \pm \sqrt{i^{2} \frac{7}{4}}$

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

$x - \frac{3}{2} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 7}{4}}$

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

$x - \frac{3}{2} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 7}{2^{2} \cdot 1}}$

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

$x - \frac{3}{2} = \pm \sqrt{i^{2} ({(\frac{1}{2})}^{2} \cdot 7)}$

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

$x - \frac{3}{2} = \pm \sqrt{{(\frac{i}{2})}^{2} \cdot 7}$

Pull terms out from under the radical.

$x - \frac{3}{2} = \pm \frac{i}{2} \sqrt{7}$

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

$x - \frac{3}{2} = \pm \frac{i \sqrt{7}}{2}$

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

Add $\frac{3}{2}$ to both sides of the equation.

$x = \pm \frac{i \sqrt{7}}{2} + \frac{3}{2}$

Reorder $\pm \frac{i \sqrt{7}}{2}$ and $\frac{3}{2}$ .

$x = \frac{3}{2} \pm \frac{i \sqrt{7}}{2}$

Solve by Completing the Square x^2-3x+4=0

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