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

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

Step 1

Add $4$ to both sides of the equation.

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

Step 2

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

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

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

Simplify the left side.

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $4$ by $2$ .

$x^{2} + x = 2$

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 $b$ .

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

Step 4

Add the term to each side of the equation.

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

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

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

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

One to any power is one.

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

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

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

Simplify the right side.

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

Simplify each term.

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

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

One to any power is one.

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

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

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

To write $2$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Combine $2$ and $\frac{4}{4}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $2$ by $4$ .

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

Add $8$ and $1$ .

$x^{2} + x + \frac{1}{4} = \frac{9}{4}$

Step 6

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

${(x + \frac{1}{2})}^{2} = \frac{9}{4}$

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.

$x + \frac{1}{2} = \pm \sqrt{\frac{9}{4}}$

Simplify $\pm \sqrt{\frac{9}{4}}$ .

Rewrite $\sqrt{\frac{9}{4}}$ as $\frac{\sqrt{9}}{\sqrt{4}}$ .

$x + \frac{1}{2} = \pm \frac{\sqrt{9}}{\sqrt{4}}$

Simplify the numerator.

Rewrite $9$ as $3^{2}$ .

$x + \frac{1}{2} = \pm \frac{\sqrt{3^{2}}}{\sqrt{4}}$

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

$x + \frac{1}{2} = \pm \frac{3}{\sqrt{4}}$

Simplify the denominator.

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

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

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

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

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

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

Move all terms not containing $x$ to the right side of the equation.

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

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

Combine the numerators over the common denominator.

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

Subtract $1$ from $3$ .

$x = \frac{2}{2}$

Divide $2$ by $2$ .

$x = 1$

Next, use the negative value of the $\pm$ to find the second solution.

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

Move all terms not containing $x$ to the right side of the equation.

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

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

Combine the numerators over the common denominator.

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

Subtract $1$ from $- 3$ .

$x = \frac{- 4}{2}$

Divide $- 4$ by $2$ .

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 2$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.