Solve by Completing the Square x^2-x=3

$x^{2} - x = 3$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{1}{2})}^{2}$

Step 2

Add the term to each side of the equation.

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

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 $- \frac{1}{2}$ .

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

Apply the product rule to $\frac{1}{2}$ .

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

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

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

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

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

One to any power is one.

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

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

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

Simplify the right side.

Simplify $3 + {(- \frac{1}{2})}^{2}$ .

Simplify each term.

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

Apply the product rule to $- \frac{1}{2}$ .

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

Apply the product rule to $\frac{1}{2}$ .

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

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

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

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

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

One to any power is one.

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

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

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

To write $3$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Combine $3$ and $\frac{4}{4}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $3$ by $4$ .

$x^{2} - x + \frac{1}{4} = \frac{12 + 1}{4}$

Add $12$ and $1$ .

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

Step 4

Factor the perfect trinomial square into ${(x - \frac{1}{2})}^{2}$ .

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

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.

$x - \frac{1}{2} = \pm \sqrt{\frac{13}{4}}$

Simplify $\pm \sqrt{\frac{13}{4}}$ .

Rewrite $\sqrt{\frac{13}{4}}$ as $\frac{\sqrt{13}}{\sqrt{4}}$ .

$x - \frac{1}{2} = \pm \frac{\sqrt{13}}{\sqrt{4}}$

Simplify the denominator.

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

$x - \frac{1}{2} = \pm \frac{\sqrt{13}}{\sqrt{2^{2}}}$

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

$x - \frac{1}{2} = \pm \frac{\sqrt{13}}{2}$

Add $\frac{1}{2}$ to both sides of the equation.

$x = \pm \frac{\sqrt{13}}{2} + \frac{1}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{13}}{2} + \frac{1}{2}$

Decimal Form:

$x = 2.30277563 \dots, - 1.30277563 \dots$

Solve by Completing the Square x^2-x=3

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