Solve by Completing the Square x^2-x-7=0

$x^{2} - x - 7 = 0$

Step 1

Add $7$ to both sides of the equation.

$x^{2} - x = 7$

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.

$x^{2} - x + {(- \frac{1}{2})}^{2} = 7 + {(- \frac{1}{2})}^{2}$

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

$x^{2} - x + {(- 1)}^{2} {(\frac{1}{2})}^{2} = 7 + {(- \frac{1}{2})}^{2}$

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

$x^{2} - x + {(- 1)}^{2} \frac{1^{2}}{2^{2}} = 7 + {(- \frac{1}{2})}^{2}$

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

$x^{2} - x + 1 \frac{1^{2}}{2^{2}} = 7 + {(- \frac{1}{2})}^{2}$

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

$x^{2} - x + \frac{1^{2}}{2^{2}} = 7 + {(- \frac{1}{2})}^{2}$

One to any power is one.

$x^{2} - x + \frac{1}{2^{2}} = 7 + {(- \frac{1}{2})}^{2}$

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

$x^{2} - x + \frac{1}{4} = 7 + {(- \frac{1}{2})}^{2}$

Simplify the right side.

Simplify $7 + {(- \frac{1}{2})}^{2}$ .

Simplify each term.

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

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

$x^{2} - x + \frac{1}{4} = 7 + {(- 1)}^{2} {(\frac{1}{2})}^{2}$

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

$x^{2} - x + \frac{1}{4} = 7 + {(- 1)}^{2} \frac{1^{2}}{2^{2}}$

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

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

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

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

One to any power is one.

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

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

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

To write $7$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x^{2} - x + \frac{1}{4} = 7 \cdot \frac{4}{4} + \frac{1}{4}$

Combine $7$ and $\frac{4}{4}$ .

$x^{2} - x + \frac{1}{4} = \frac{7 \cdot 4}{4} + \frac{1}{4}$

Combine the numerators over the common denominator.

$x^{2} - x + \frac{1}{4} = \frac{7 \cdot 4 + 1}{4}$

Simplify the numerator.

Multiply $7$ by $4$ .

$x^{2} - x + \frac{1}{4} = \frac{28 + 1}{4}$

Add $28$ and $1$ .

$x^{2} - x + \frac{1}{4} = \frac{29}{4}$

Step 5

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

${(x - \frac{1}{2})}^{2} = \frac{29}{4}$

Step 6

Solve the equation for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x - \frac{1}{2} = \pm \sqrt{\frac{29}{4}}$

Simplify $\pm \sqrt{\frac{29}{4}}$ .

Rewrite $\sqrt{\frac{29}{4}}$ as $\frac{\sqrt{29}}{\sqrt{4}}$ .

$x - \frac{1}{2} = \pm \frac{\sqrt{29}}{\sqrt{4}}$

Simplify the denominator.

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

$x - \frac{1}{2} = \pm \frac{\sqrt{29}}{\sqrt{2^{2}}}$

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

$x - \frac{1}{2} = \pm \frac{\sqrt{29}}{2}$

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

$x = \pm \frac{\sqrt{29}}{2} + \frac{1}{2}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{29}}{2} + \frac{1}{2}$

Decimal Form:

$x = 3.19258240 \dots, - 2.19258240 \dots$

Solve by Completing the Square x^2-x-7=0

Related Questions

Support LCMGCF.com by disabling your adblocker.