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

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

Step 1

Add $1$ to both sides of the equation.

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

Step 2

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

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 the left side.

Simplify each term.

Cancel the common factor of $2$ .

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 $2$ .

Factor $2$ out of $- 4 x$ .

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

Cancel the common factors.

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 $1$ .

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

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 $b$ .

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

Step 4

Add the term to each side of the equation.

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

Step 5

Simplify the equation.

Simplify the left side.

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

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

Simplify the right side.

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

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

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

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 $2$ .

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

Step 6

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

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

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.

$x - 1 = \pm \sqrt{\frac{3}{2}}$

Simplify $\pm \sqrt{\frac{3}{2}}$ .

Rewrite $\sqrt{\frac{3}{2}}$ as $\frac{\sqrt{3}}{\sqrt{2}}$ .

$x - 1 = \pm \frac{\sqrt{3}}{\sqrt{2}}$

Multiply $\frac{\sqrt{3}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$x - 1 = \pm \frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{3}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$x - 1 = \pm \frac{\sqrt{3} \sqrt{2}}{\sqrt{2} \sqrt{2}}$

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

$x - 1 = \pm \frac{\sqrt{3} \sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

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

$x - 1 = \pm \frac{\sqrt{3} \sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

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

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

Add $1$ and $1$ .

$x - 1 = \pm \frac{\sqrt{3} \sqrt{2}}{{\sqrt{2}}^{2}}$

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

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

$x - 1 = \pm \frac{\sqrt{3} \sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

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

$x - 1 = \pm \frac{\sqrt{3} \sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

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

$x - 1 = \pm \frac{\sqrt{3} \sqrt{2}}{2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x - 1 = \pm \frac{\sqrt{3} \sqrt{2}}{2^{\frac{2}{2}}}$

Rewrite the expression.

$x - 1 = \pm \frac{\sqrt{3} \sqrt{2}}{2^{1}}$

Evaluate the exponent.

$x - 1 = \pm \frac{\sqrt{3} \sqrt{2}}{2}$

Simplify the numerator.

Combine using the product rule for radicals.

$x - 1 = \pm \frac{\sqrt{3 \cdot 2}}{2}$

Multiply $3$ by $2$ .

$x - 1 = \pm \frac{\sqrt{6}}{2}$

Add $1$ to both sides of the equation.

$x = \pm \frac{\sqrt{6}}{2} + 1$

Step 8

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{6}}{2} + 1$

Decimal Form:

$x = 2.22474487 \dots, - 0.22474487 \dots$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.