Solve by Completing the Square x^2-x-1=0

$x^{2} - x - 1 = 0$

Step 1

Add $1$ to both sides of the equation.

$x^{2} - x = 1$

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 $b$ .

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

Step 3

Add the term to each side of the equation.

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

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

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

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

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

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

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

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

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

One to any power is one.

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

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

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

Simplify the right side.

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

Simplify each term.

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

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

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

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

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

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

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

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

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

One to any power is one.

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

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

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

Simplify the expression.

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Add $4$ and $1$ .

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

Step 5

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

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

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.

$x - \frac{1}{2} = \pm \sqrt{\frac{5}{4}}$

Simplify $\pm \sqrt{\frac{5}{4}}$ .

Rewrite $\sqrt{\frac{5}{4}}$ as $\frac{\sqrt{5}}{\sqrt{4}}$ .

$x - \frac{1}{2} = \pm \frac{\sqrt{5}}{\sqrt{4}}$

Simplify the denominator.

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

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

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

$x - \frac{1}{2} = \pm \frac{\sqrt{5}}{2}$

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

$x = \pm \frac{\sqrt{5}}{2} + \frac{1}{2}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{5}}{2} + \frac{1}{2}$

Decimal Form:

$x = 1.61803398 \dots, - 0.61803398 \dots$

Solve by Completing the Square x^2-x-1=0

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