Solve by Completing the Square 2x^2-x-6=0

$2 x^{2} - x - 6 = 0$

Step 1

Add $6$ to both sides of the equation.

$2 x^{2} - x = 6$

Step 2

Divide each term in $2 x^{2} - x = 6$ by $2$ and simplify.

Divide each term in $2 x^{2} - x = 6$ by $2$ .

$\frac{2 x^{2}}{2} + \frac{- x}{2} = \frac{6}{2}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} + \frac{- x}{2} = \frac{6}{2}$

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

$x^{2} + \frac{- x}{2} = \frac{6}{2}$

Move the negative in front of the fraction.

$x^{2} - \frac{x}{2} = \frac{6}{2}$

Simplify the right side.

Divide $6$ by $2$ .

$x^{2} - \frac{x}{2} = 3$

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.

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

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

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

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

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

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

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

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

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

One to any power is one.

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

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

$x^{2} - \frac{x}{2} + \frac{1}{16} = 3 + {(- \frac{1}{4})}^{2}$

Simplify the right side.

Simplify $3 + {(- \frac{1}{4})}^{2}$ .

Simplify each term.

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

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

$x^{2} - \frac{x}{2} + \frac{1}{16} = 3 + {(- 1)}^{2} {(\frac{1}{4})}^{2}$

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

$x^{2} - \frac{x}{2} + \frac{1}{16} = 3 + {(- 1)}^{2} \frac{1^{2}}{4^{2}}$

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

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

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

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

One to any power is one.

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

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

$x^{2} - \frac{x}{2} + \frac{1}{16} = 3 + \frac{1}{16}$

To write $3$ as a fraction with a common denominator, multiply by $\frac{16}{16}$ .

$x^{2} - \frac{x}{2} + \frac{1}{16} = 3 \cdot \frac{16}{16} + \frac{1}{16}$

Combine $3$ and $\frac{16}{16}$ .

$x^{2} - \frac{x}{2} + \frac{1}{16} = \frac{3 \cdot 16}{16} + \frac{1}{16}$

Combine the numerators over the common denominator.

$x^{2} - \frac{x}{2} + \frac{1}{16} = \frac{3 \cdot 16 + 1}{16}$

Simplify the numerator.

Multiply $3$ by $16$ .

$x^{2} - \frac{x}{2} + \frac{1}{16} = \frac{48 + 1}{16}$

Add $48$ and $1$ .

$x^{2} - \frac{x}{2} + \frac{1}{16} = \frac{49}{16}$

Step 6

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

${(x - \frac{1}{4})}^{2} = \frac{49}{16}$

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.

$x - \frac{1}{4} = \pm \sqrt{\frac{49}{16}}$

Simplify $\pm \sqrt{\frac{49}{16}}$ .

Rewrite $\sqrt{\frac{49}{16}}$ as $\frac{\sqrt{49}}{\sqrt{16}}$ .

$x - \frac{1}{4} = \pm \frac{\sqrt{49}}{\sqrt{16}}$

Simplify the numerator.

Rewrite $49$ as $7^{2}$ .

$x - \frac{1}{4} = \pm \frac{\sqrt{7^{2}}}{\sqrt{16}}$

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

$x - \frac{1}{4} = \pm \frac{7}{\sqrt{16}}$

Simplify the denominator.

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

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

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

$x - \frac{1}{4} = \pm \frac{7}{4}$

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.

$x - \frac{1}{4} = \frac{7}{4}$

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

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

$x = \frac{7}{4} + \frac{1}{4}$

Combine the numerators over the common denominator.

$x = \frac{7 + 1}{4}$

Add $7$ and $1$ .

$x = \frac{8}{4}$

Divide $8$ by $4$ .

$x = 2$

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

$x - \frac{1}{4} = - \frac{7}{4}$

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

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

$x = - \frac{7}{4} + \frac{1}{4}$

Combine the numerators over the common denominator.

$x = \frac{- 7 + 1}{4}$

Add $- 7$ and $1$ .

$x = \frac{- 6}{4}$

Cancel the common factor of $- 6$ and $4$ .

Factor $2$ out of $- 6$ .

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

Cancel the common factors.

Factor $2$ out of $4$ .

$x = \frac{2 \cdot - 3}{2 \cdot 2}$

Cancel the common factor.

$x = \frac{2 \cdot - 3}{2 \cdot 2}$

Rewrite the expression.

$x = \frac{- 3}{2}$

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

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

$x = 2, - \frac{3}{2}$

Solve by Completing the Square 2x^2-x-6=0

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