Solve by Completing the Square x^2+x-8=0

$x^{2} + x - 8 = 0$

Step 1

Add $8$ to both sides of the equation.

$x^{2} + x = 8$

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.

$x^{2} + x + {(\frac{1}{2})}^{2} = 8 + {(\frac{1}{2})}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Simplify each term.

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

$x^{2} + x + \frac{1^{2}}{2^{2}} = 8 + {(\frac{1}{2})}^{2}$

One to any power is one.

$x^{2} + x + \frac{1}{2^{2}} = 8 + {(\frac{1}{2})}^{2}$

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

$x^{2} + x + \frac{1}{4} = 8 + {(\frac{1}{2})}^{2}$

Simplify the right side.

Simplify $8 + {(\frac{1}{2})}^{2}$ .

Simplify each term.

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

$x^{2} + x + \frac{1}{4} = 8 + \frac{1^{2}}{2^{2}}$

One to any power is one.

$x^{2} + x + \frac{1}{4} = 8 + \frac{1}{2^{2}}$

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

$x^{2} + x + \frac{1}{4} = 8 + \frac{1}{4}$

To write $8$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x^{2} + x + \frac{1}{4} = 8 \cdot \frac{4}{4} + \frac{1}{4}$

Combine $8$ and $\frac{4}{4}$ .

$x^{2} + x + \frac{1}{4} = \frac{8 \cdot 4}{4} + \frac{1}{4}$

Combine the numerators over the common denominator.

$x^{2} + x + \frac{1}{4} = \frac{8 \cdot 4 + 1}{4}$

Simplify the numerator.

Multiply $8$ by $4$ .

$x^{2} + x + \frac{1}{4} = \frac{32 + 1}{4}$

Add $32$ and $1$ .

$x^{2} + x + \frac{1}{4} = \frac{33}{4}$

Step 5

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

${(x + \frac{1}{2})}^{2} = \frac{33}{4}$

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.

$x + \frac{1}{2} = \pm \sqrt{\frac{33}{4}}$

Simplify $\pm \sqrt{\frac{33}{4}}$ .

Rewrite $\sqrt{\frac{33}{4}}$ as $\frac{\sqrt{33}}{\sqrt{4}}$ .

$x + \frac{1}{2} = \pm \frac{\sqrt{33}}{\sqrt{4}}$

Simplify the denominator.

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

$x + \frac{1}{2} = \pm \frac{\sqrt{33}}{\sqrt{2^{2}}}$

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

$x + \frac{1}{2} = \pm \frac{\sqrt{33}}{2}$

Subtract $\frac{1}{2}$ from both sides of the equation.

$x = \pm \frac{\sqrt{33}}{2} - \frac{1}{2}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{33}}{2} - \frac{1}{2}$

Decimal Form:

$x = 2.37228132 \dots, - 3.37228132 \dots$

Solve by Completing the Square x^2+x-8=0

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