Solve by Completing the Square x^2+22x=31

$x^{2} + 22 x = 31$

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(11)}^{2}$

Add the term to each side of the equation.

$x^{2} + 22 x + {(11)}^{2} = 31 + {(11)}^{2}$

Simplify the equation.

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Raise $11$ to the power of $2$ .

$x^{2} + 22 x + 121 = 31 + {(11)}^{2}$

Simplify $31 + {(11)}^{2}$ .

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Raise $11$ to the power of $2$ .

$x^{2} + 22 x + 121 = 31 + 121$

Add $31$ and $121$ .

$x^{2} + 22 x + 121 = 152$

Factor the perfect trinomial square into ${(x + 11)}^{2}$ .

${(x + 11)}^{2} = 152$

Solve the equation for $x$ .

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Take the square root of each side of the equation to set up the solution for $x$

${(x + 11)}^{2 \cdot \frac{1}{2}} = \pm \sqrt{152}$

Remove the perfect root factor $x + 11$ under the radical to solve for $x$ .

$x + 11 = \pm \sqrt{152}$

Simplify the right side of the equation.

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Rewrite $152$ as $2^{2} \cdot 38$ .

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Factor $4$ out of $152$ .

$x + 11 = \pm \sqrt{4 (38)}$

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

$x + 11 = \pm \sqrt{2^{2} \cdot 38}$

Pull terms out from under the radical.

$x + 11 = \pm 2 \sqrt{38}$

Remove parentheses.

$x + 11 = \pm 2 \sqrt{38}$

Subtract $11$ from both sides of the equation.

$x = \pm 2 \sqrt{38} - 11$

The result can be shown in multiple forms.

Exact Form:

$x = \pm 2 \sqrt{38} - 11$

Decimal Form:

$x = 1.32882800 \dots, - 23.32882800 \dots$

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