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

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

Step 1

Add $1$ to both sides of the equation.

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

Step 2

Divide each term in $2 x^{2} - x = 1$ by $2$ and simplify.

Divide each term in $2 x^{2} - x = 1$ by $2$ .

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

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

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

Move the negative in front of the fraction.

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

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 $b$ .

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

Step 4

Add the term to each side of the equation.

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

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 $- \frac{1}{4}$ .

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

Apply the product rule to $\frac{1}{4}$ .

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

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

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

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

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

One to any power is one.

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

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

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

Simplify the right side.

Simplify $\frac{1}{2} + {(- \frac{1}{4})}^{2}$ .

Simplify each term.

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

Apply the product rule to $- \frac{1}{4}$ .

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

Apply the product rule to $\frac{1}{4}$ .

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

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

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

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

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

One to any power is one.

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

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

$x^{2} - \frac{x}{2} + \frac{1}{16} = \frac{1}{2} + \frac{1}{16}$

To write $\frac{1}{2}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$x^{2} - \frac{x}{2} + \frac{1}{16} = \frac{1}{2} \cdot \frac{8}{8} + \frac{1}{16}$

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

Multiply $\frac{1}{2}$ by $\frac{8}{8}$ .

$x^{2} - \frac{x}{2} + \frac{1}{16} = \frac{8}{2 \cdot 8} + \frac{1}{16}$

Multiply $2$ by $8$ .

$x^{2} - \frac{x}{2} + \frac{1}{16} = \frac{8}{16} + \frac{1}{16}$

Combine the numerators over the common denominator.

$x^{2} - \frac{x}{2} + \frac{1}{16} = \frac{8 + 1}{16}$

Add $8$ and $1$ .

$x^{2} - \frac{x}{2} + \frac{1}{16} = \frac{9}{16}$

Step 6

Factor the perfect trinomial square into ${(x - \frac{1}{4})}^{2}$ .

${(x - \frac{1}{4})}^{2} = \frac{9}{16}$

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.

$x - \frac{1}{4} = \pm \sqrt{\frac{9}{16}}$

Simplify $\pm \sqrt{\frac{9}{16}}$ .

Rewrite $\sqrt{\frac{9}{16}}$ as $\frac{\sqrt{9}}{\sqrt{16}}$ .

$x - \frac{1}{4} = \pm \frac{\sqrt{9}}{\sqrt{16}}$

Simplify the numerator.

Rewrite $9$ as $3^{2}$ .

$x - \frac{1}{4} = \pm \frac{\sqrt{3^{2}}}{\sqrt{16}}$

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

$x - \frac{1}{4} = \pm \frac{3}{\sqrt{16}}$

Simplify the denominator.

Rewrite $16$ as $4^{2}$ .

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

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

$x - \frac{1}{4} = \pm \frac{3}{4}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x - \frac{1}{4} = \frac{3}{4}$

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

Add $\frac{1}{4}$ to both sides of the equation.

$x = \frac{3}{4} + \frac{1}{4}$

Combine the numerators over the common denominator.

$x = \frac{3 + 1}{4}$

Add $3$ and $1$ .

$x = \frac{4}{4}$

Divide $4$ by $4$ .

$x = 1$

Next, use the negative value of the $\pm$ to find the second solution.

$x - \frac{1}{4} = - \frac{3}{4}$

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

Add $\frac{1}{4}$ to both sides of the equation.

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

Combine the numerators over the common denominator.

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

Add $- 3$ and $1$ .

$x = \frac{- 2}{4}$

Cancel the common factor of $- 2$ and $4$ .

Factor $2$ out of $- 2$ .

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

Cancel the common factors.

Factor $2$ out of $4$ .

$x = \frac{2 \cdot - 1}{2 \cdot 2}$

Cancel the common factor.

$x = \frac{2 \cdot - 1}{2 \cdot 2}$

Rewrite the expression.

$x = \frac{- 1}{2}$

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - \frac{1}{2}$

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

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