Solve by Completing the Square 2x^2+4x+1=0

$2 x^{2} + 4 x + 1 = 0$

Subtract $1$ from both sides of the equation.

$2 x^{2} + 4 x = - 1$

Divide each term by $2$ and simplify.

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Divide each term in $2 x^{2} + 4 x = - 1$ by $2$ .

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

Simplify each term.

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Cancel the common factor of $2$ .

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Cancel the common factor.

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

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

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

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

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Factor $2$ out of $4 x$ .

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

Cancel the common factors.

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Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $2 x$ by $1$ .

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

Move the negative in front of the fraction.

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

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

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

Add the term to each side of the equation.

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

Simplify the equation.

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One to any power is one.

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

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

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One to any power is one.

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Add $- 1$ and $2$ .

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

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

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

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 $x$

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

Remove the perfect root factor $x + 1$ under the radical to solve for $x$ .

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

Simplify the right side of the equation.

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Rewrite $\sqrt{\frac{1}{2}}$ as $\frac{\sqrt{1}}{\sqrt{2}}$ .

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

Any root of $1$ is $1$ .

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

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

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

Combine and simplify the denominator.

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Multiply $\frac{1}{\sqrt{2}}$ and $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

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

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $1$ by $1$ .

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

Evaluate the exponent.

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

Remove parentheses.

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

Subtract $1$ from both sides of the equation.

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 0.29289321 \dots, - 1.70710678 \dots$

Solve by Completing the Square 2x^2+4x+1=0

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